## 1. Intro

In this unit you will learn about measures of position, or location, within a data set. One important measure of position, which will be used extensively later in the course, tells us the position of a data value relative to the standard deviation. Other measures tell us location in terms of groups (or percents) of the data set.

Once you have created a distribution of your data, you can use its shape, center, and spread to tell the story of your underlying data.

The most important idea that you need to take from this unit is that of a probability density curve, the graphical representation of a continuous random variable. When you are looking at a histogram of continuous data, you can almost imagine a smooth curve making the same shape as the histogram’s bars. For example, if you think back to the example from Unit 1 concerning a state’s residents living in poverty, we produced the following histogram:

A smooth curve that has (roughly) the same shape as this histogram would be something like:

The smooth curve that represents our histogram is called a density curve and it has some cool properties. First, it is always on or above our horizontal axis. Since our vertical axis represents a count or percentage of data falling in a particular class, there can’t be a negative amount of data in a class. The other property is that the total area under an entire density curve is 1 (or 100%). Since a density curve represents our data, ALL of our data…or 100% of it…must be included in the distribution. We’re going to routinely utilize the result that an area (or region) under a density curve represents the probability of obtaining results falling in that area. So remember, AREA = PERCENTAGE OR PROBABILITY. Keep reminding yourself: AREA = PERCENTAGE OR PROBABILITY.

Again, the main purpose of a density function is to be a smooth and continuous representation of our actual data. Because the density function is a “model” of our data, we will use Greek letters such as μ and σ to represent the mean and standard deviation of the density curve. Statistics is full of symbols; it is most important to remember that x  and s represent the mean and standard deviation, respectively, of a SAMPLE, while μ and σ represent the mean and standard deviation, respectively, of a POPULATION. The density curve is a stand-in for our population.

To begin, let’s investigate the distributions of two continuous random variables, the uniform distribution and the normal distribution, which will be the focus of our statistical studies from here on out.

## 2. Uniform Distribution

One simple, basic example of a continuous random variable is one where the random variable X can take any value in a given interval with an equally likely probability. The distribution of such a random variable is the uniform distribution.

Image you show up for work one morning and are told there will be a fire alarm drill sometime during the eight-hour day. Fire drills don’t make sense if everyone knows when the drill will take place, so all you know is that sometime during the day, a drill will take place. This means that at every moment there is an equally likely chance that the fire drill will take place. Together with the information that the drill will happen, i.e., there is a 100% = 1 probability that it will occur, we get the following distribution:

Why is the probability fixed at 1/8? Use the facts that (1) there are 8 hours during which the drill can take place, and (2) there is a 100% probability of the drill occurring. Since the uniform distribution is a rectangle, and the area of any rectangle is A(length× (width), we get:

1 = 8 × height

and solving for height gives us:

Now you can determine the probabilities of the drill taking place during any time interval you choose. For example, the probability that the drill will occur during your lunch hour (from 12:00 p.m. to 1:00 p.m.) is simply the area of the region shown in red:

Once again, remember that AREA = PERCENTAGE OR PROBABILITY. The discussion of probability density curves always starts with the uniform distribution, because everyone knows how to calculate areas of rectangles. And it’s easy to see how the concepts of area and probability are linked.

## 3. The Normal Distribution

For the majority of the remainder of this class, we’ll be focusing on variables that have a (roughly) normal distribution. For example, data sets consisting of physical measurements (heights, weights, lengths of bones, and so on) for adults of the same species and sex often follow a similar pattern: most individuals are clumped around the average or mean of the population, with numbers decreasing the farther values are from the average in either direction.

The shape of any normal curve is a single-peaked, symmetric distribution that is bell-shaped. A normally distributed random variable, or a variable with a normal probability distribution, is a continuous random variable that has a relative frequency histogram in the shape of a normal curve. This curve is also called the normal density curve. The actual functional notation for creating the normal curve is quite complex:

where μ and σ are the mean and standard deviation of the population of data.

What this formula tells us is that any mean μ and standard deviation σ completely define a unique normal curve. Recall that μ tells us the “center” of the peak while σ describes the overall “fatness” of the data set. A small σ value indicates a tall, skinny data set, while a larger value of σ results in a shorter, more spread out data set. Each normal distribution is indicated by the symbols N(μ,σ) . For example, the normal distribution N(0,1) is called the standard normal distribution, and it has a mean of 0 and a standard deviation of 1.

Properties of a Normal Distribution

1. A normal distribution is bell-shaped and symmetric about its mean.
2. A normal distribution is completely defined by its mean, µ, and standard deviation, σ.
3. The total area under a normal distribution curve equals 1.
4. The x-axis is a horizontal asymptote for a normal distribution curve.

A graphical representation of the Normal Distribution curve below:

Because there are an infinite number of possibilities for µ and σ, there are an infinite number of normal curves. In order to determine probabilities for each normally distributed random variable, we would have to perform separate probability calculations for each normal distribution.

One amazing fact about any normal distribution is called the 68-95-99.7 Rule, or more concisely, the empirical rule. This rule states that:

• Roughly 68% of all data observations fall within one standard deviation on either side of the mean. Thus, there is a 68% chance of a variable having a value within one standard deviation of the mean
• Roughly 95% of all data observations fall within two standard deviations on either side of the mean. Thus, there is a 95% chance of a variable having a value within two standard deviations of the mean
• Roughly 99.7% of all data observations fall within three standard deviations on either side of the mean. Thus, there is a 99.7% chance of a variable having a value within three standard deviations of the mean

A graphical representation of the empirical rule is shown in the following figure:

##### Example:

Suppose a variable has mean μ = 17   and standard deviation σ = 3.4. Then, according to the empirical rule:

• Approximately 68% of individual data values will lie between: 17 – 3.4 = 13.6 and 17 + 3.4 = 20.4. In interval notation we write: (13.6, 20.4).
• Approximately 95% of individual data values will lie between 17 – 2⋅3.4 = 10.2 and 17 + 2⋅3.4 = 23.8. In interval notation we write: (10.2, 23.8).
• Approximately 99.7% of individual data values will lie between 17 – 3⋅3.4 = 6.8 and 17 + 3⋅3.4 = 27.2. In interval notation we write: (6.8, 27.2).

The results from the third bullet point illustrate how a data value of, say, 2.1 (which is less than 6.8) or a data value of, say, 33.2 (a value greater than 27.2) would both be very unusual, since almost all data values should lie between 6.8 and 27.2.

#### Back to the Standard Normal Curve

All normal distributions, regardless of their mean and standard deviation, share the Empirical Rule. With some very simple mathematics, we can “transform” any normal distribution into the standard normal distribution. This is called a z-transform.

Using the z-transformation, any data set that is normally distributed can be converted to the same standard normal distribution by the conversion:

where X is the normally distributed random variable, and Z is a random variable following the standard normal distribution.

Notice when X = μ that Z = (μ – μ)/σ = 0, which explains how Z transforms our mean to 0.

Properties of the Standard Normal Distribution

1. The standard normal distribution is bell-shaped and symmetric about its mean.
2. The standard normal distribution is completely defined by its mean, µ = 0, and standard deviation,  σ = 1.
3. The total area under the standard normal distribution curve equals 1.
4. The x-axis is a horizontal asymptote for the standard normal distribution curve.

## 4. The z-Score

Given any data value, we can identify how far that data value is away from the mean, simply by doing a subtraction x – μ. This value will be positive if your data value lies above (to the right) of the mean, and negative if it lies below (to the left) of the mean. But what we’d really like to know is, relative to the spread of our data set, how far is x from μ? Remember that the standard deviation σ gives us a measure of how spread out our entire set of individual data values is.

The z-score for any single data value can be found by the formula (in English):

or with symbols (as seen before!):

Obviously a z-score will be positive if the data value lies above (to the right) of the mean, and negative if the data value lies below (to the left) of the mean.

Example 6.1: Calculating and Graphing z-Values

Given a normal distribution with μ = 48 and s = 5, convert an x-value of 45 to a z-value and indicate where this z-value would be on the standard normal distribution.

Solution

Begin by finding the z-score for x = 45 as follows.

Now draw each of the distributions, marking a standard score of z = −0.60 on the standard normal distribution.

The distribution on the left is a normal distribution with a mean of 48 and a standard deviation of 5. The distribution on the right is a standard normal distribution with a standard score of z = −0.60 indicated.

Z-scores measure the distance of any data point from the mean in units of standard deviations and are useful because they allow us to compare the relative positions of data values in different samples. In other words, the z-score allows us to standardize two or more normal distributions, or more appropriately, to put them on the same scale. Therefore, we’ll be able to compare relative positions of data values within their own distribution to determine which data values are closer to or farther from the mean. A prime example for this is to compare the test scores for two students, one who scored a 28 on the ACT (scores range from 1 – 36) and another who scored a 1280 on the SAT (scores range from 400 – 1600). Who, relative to their associated exam, scored better?

##### Example

Your statistics exam score was 0.67 standard deviations better than the class average; your biology score was 0.7 standard deviations better than the class average; your kayaking score was only 0.5 standard deviations better than the class average.  Therefore, even though your actual score on the biology exam was the lowest of the three exam scores, relative to the distribution of all class exam scores, your biology exam score was the highest relative grade.

#### Finding an Area (Proportion) Given a Specific Z-Value

To determine the area under the N(0, 1) curve for any data value that does not fall exactly 1, 2, or 3 standard deviations above or below the mean actually requires some calculus. Lucky for us, areas under the N(0, 1) curve can be obtained in numerous other ways, including technology (TI-83/84, Excel) and a table of values. Search the Internet for “standard normal table” and you’ll find hundreds of tables illustrating z-scores and their associated areas. The majority of these methods report the area to the left of the specified z-score z, no matter where it lies. This comes from a calculus operation of integration, which finds an area from the start of a distribution (i.e., the far left-tail) up to the z-score. Two images are provided.

There are three types of area calculations that you will be performing, each requiring slightly different work:

• For areas to the left of z: simply use the area provided by a table or technology.
• For areas to the right of z: because the total area under a density curve is 1 (100%), simply calculate: 1 − area to the left of z0.
• For areas between two z-values, say zand z1 (where z< z1): find the area to the left of z1 and subtract from it the area to the left of z0.

#### Finding a Z-Value Given an Area

This is a slightly more challenging task than calculating an area, because you basically work “backwards” from an algebraic standpoint. It’s important to realize that a Standard Normal Table has two parts: (1) the top and side margins, which form the tenths and hundredths of a z-score, and (2) the body of the table, which are all the area (probability) values. Also, remember that the Standard Normal Table only provides us information on the area (probability) to the left of a z-score. A small excerpt of Table B from Appendix A is shown below.

Notice that the z-values given in the table are rounded to two decimal places. The first decimal place of each  z-value is listed in the left column, with the second decimal place in the top row. Where the appropriate row and column intersect, we find the amount of area under the standard normal curve to the left of that particular z-value.

Example : Finding Area to the Left of a Positive z-Value Using a Cumulative Normal Table

Find the area under the standard normal curve to the left of z = 1.37.

Solution

To read the table, we must break the given z-value (1.37) into two parts: one containing the first decimal place (1.3) and the other containing the second decimal place (0.07). So, in Table B from Appendix A, look across the row labeled 1.3 and down the column labeled 0.07. The row and column intersect at 0.9147. Thus, the area under the standard normal curve to the left of z = 1.37 is 0.9147.

Using a TI-83/84 Plus calculator, we can find a value of the area to the left of a z-score. To obtain the solution using a TI-83/84 Plus calculator, perform the following steps.

• Press 2nd and then Vars to access the DISTR menu.
• Choose option 2:normalcdf( .
• Enter lower bound, upper bound, µ , σ. Note If you want to find area under the standard normal curve, as in this example, then you do not need to enter µ or σ.
• Since we are asked to find the area to the left of z, the lower bound is -∞. From the empirical rule we know that after about 3 standard deviations away from the mean we have accounted for almost all of the data, so for our lower bound we will simply use a very negative number.We cannot enter -∞ into the calculator, so we will enter a very small value for the lower endpoint, such as -1099. This number appears as -1E99 when entered correctly into the calculator. To enter -1E99, press(-) 1 [2nd][ , ]99. This appears on the screen as normcdf(-1E99,1.37,0,1).

If we are given an area (or probability) value, we need to first locate it in the body of a table, then track our way up and to the left in order to piece together the z-score that relates to the specified area. Keep in mind that you may not find the exact area value in the body of the table…so just use the closest value you can find, and then identify the proper z-score.

One calculation that will be used frequently in the coming chapters is to identify the two z-scores that separate a specific area in the middle of the standard normal distribution.

##### Example

Suppose we want to know which two z-scores separate out the middle 95% of the data. From the empirical rule, we already know the z-scores that do this are ±2 (2 standard deviations on either side of the mean). In reality, it’s not exactly ±2, but close enough for rough calculations.

To find the exact two z-scores, we use the following logic: If the middle portion is 95% = 0.95, then how much area lies outside of the middle (to the left and right)? A simple subtraction solves this! 1 – 0.95 = 0.05. The “outside” area, 0.05, must be split equally between the two tails (because of symmetry!). Therefore, dividing 0.05 by two gives us an area of 0.025 in each tail.

Using a standard normal table “backwards,” we first look through the body of the table to find an area closest to 0.025. The z-score corresponding to a left-tail area of 0.025 is z = −1.96. Now, therefore, the upper z-score will be z = 1.96, by the symmetry property of the standard normal distribution. You could also discover the upper z-score by looking up the area/probability value 0.025 + 0.95 = 0.975 in the body of the table and finding the associated z-value. By the end of the class, you will be extremely familiar with z-scores that define a central 90% (z = ± 1.645), 95% (z = ± 1.96), and 99% (z = ± 2.576).

##### Example: Find and interpret the probability of a random Normal variable

Suppose you just purchased a 2005 Honda Insight with automatic transmission. Using www.fueleconomy.gov you determine for the 2005 Honda Insights have mean highway gas milage is 56 miles per gallon with a standard deviation of 3.2. The distribution of this data has a bell-shape and is normal. You want to know the following:

a) How likely is it that your Honda Insight with automatic transition will get better than 60 miles per gallon on the highway.

b) How likely is it that your Honda Insight with automatic transition will get less than 50 miles per gallon on the highway.

c) How likely is it that your Honda Insight with aoutomatic transition will get between 52 and 62 miles per gallon on the highway.

Solution

This problem deals with data that is normally distributed with mean 56 and standard deviation 3.2, i.e., .

(a)

In symbols, we are asked to calculate P(X > 60). Sketching a normal curve and shading the area corresponding to greater than 60, gives us the graph shown. In order to calculate the appropriate area in the upper (right) tail, we must first convert our data to the standard normal distribution. The z-score for x = 60 is:

This means that 60 is 1.25 standard deviations above the mean. Notice how lining the two normal curves up as shown illustrates how the two areas are the same: P(X > 60) = P(Z > 1.25).

Using z = 1.25, we go to Table IV (or use normcdf(1.25,1E99,0,1))  to find the area to the left of z = 1.25 is 0.8943. Since we need the area to the right, we simply take 1 – 0.8943 = 0.1057.

Therefore, P(X > 60) = 0.1057 = 10.57%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 10.57% will get over 60 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 10.57% chance that your car will get over 60 miles per gallon on the highway.

(b)

In symbols, we are asked to calculate P(X < 50). Sketching a normal curve N(56, 3.2)and shading the area corresponding to less than 50, gives us the graph shown to the right.

In order to calculate the appropriate area in the lower (left) tail, we must first convert our data to the standard normal distribution. The z-score for x = 50 is:

Thus, the value 50 MPG is 1.88 standard deviations below the mean. In symbols we see: P(X < 50) = P(Z < −1.88).

Using z = -1.88, we go to Table IV (or use normcdf(-1E99,-1.88,0,1)) to find the area to the left of z = -1.88 is 0.0301. Therefore, P(X < 50) = 0.0301 = 3.01%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 3.01% will get less than 50 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 3.01% chance that your car will get less than 50 miles per gallon on the highway.

(c)

In symbols, we are asked to calculate P(58 < X < 62). Sketching a normal curve N(56, 3.2)  and shading the area corresponding to greater than 58 but less than 62, gives us the graph shown. In order to calculate the appropriate area, we must first convert both data to the standard normal distribution.

The z-score for = 58 is:

and the z-score for x = 62 is:

In terms of probability, we can now say: P(58 < X < 62) = P(0.63 < Z < 1.88).

Using z = 1.88, we go to Table IV (or use technology) to find the area to the left of z = 1.88 is 0.9699. Now, we need to remove (subtract) the area left of z = 0.63, which is 0.7357. Therefore, P(58 < X < 62) = 0.9699 – 0.7357 = 0.2342, or 23.42%. There are a couple ways to interpret this answer:

• Of all the model year 2005 Honda Insight cars produced with an automatic transmission, 23.42% will get between 58 and 62 miles per gallon on the highway.
• If you went to a car lot and purchased a new model year 2005 Honda Insight cars produced with an automatic transmission, there is a 23.42% chance that your car will get between 58 and 62 miles per gallon on the highway.

This calculation can be done with both normcdf(0.63,1.88,0,1) and normcdf(58,62,56,3.2), which will be the same.

#### Find the Value of a Random Variable Knowing a Probability Value

In these types of problems, we need to work “backwards.” Starting with a specified probability, find the specified z-score, then work our way back to the random variable. The tables of standard normal values are not a “one-way” tool! What do we mean by that? So far you’ve started with a value for a random variable (like a gas mileage value in the previous problem), turned it into a z-score, and then looked up the associated probability value for that z-score. We can use this table to work backwards! We can start with a known probability value in the body of a table, identify the z-score corresponding to that area by moving your fingers to the associated row and column, the reverse the algebra transformation from a z-score to a random variable.

If this sounds confusing, think back to the steps we took in the preceding example:

If, however, we are given an area/probability, then to work our way back to the original data value, we must first identify the appropriate z-score, and then “un-standardize” the z-score to arrive (finally!) back at the data value. How do we algebraically “undo” the z-score? Easy…just solve for the data value X:

Multiply both sides by σ to remove it from the denominator on the left side:

X – μ = Z⋅σ

Finally, add the value of μ to both sides to isolate the value of the random variable X:

X = Z⋅σ + μ

##### Example: Finding the value of a normal random variable

Instead you want to know a gas mileage for a particular probability. Find what gas mileage for your 2005 Honda Insight will get better gas mileage than 97% of all other 2005 Honda Insights with automatics transmission.

Solution

This problem again deals with data that is normally distributed with mean 56 and standard deviation 3.2, i.e., N(56, 3.2).

To find the 97% percentile gas mileage, we need to find the specific miles per gallon X that separates the bottom 97% of all gas mileages from the top 3%. So for this problem we are given a percentage/area. Sketching the normal curve gives the graph shown.

Using Table IV, we find 0.97 in the body of the table, and then identify the z-score of 1.88. Notice that the exact area 0.97 is not in the table, but the closest area of 0.9699 has the z-score of 1.88. Now we un-standardize the z-score of 1.88. In English this means we need to identify the specific gas mileage that is 1.88 standard deviations above the mean of 56. Solving for X in the Z transform gives:

Therefore, if your 2005 Honda Insight cars with an automatic transmission gets 62 mpg, it gets better miles per gallon than 97% of all 2005 Honda Insight cars with an automatic transmission.

## 5. Percentiles

Remember that the median of a data set divides the lower 50% of the data from the upper 50%. We say that the median is the 50th percentile of the data set. If a number divides the lower 34% of the data from the upper 66%, that number is the 34th percentile. In general, a kth percentile of a data set is a value that divides the data set into the lower kth percentile and the upper (1 − kth) percentile.

The computation of the kth percentile is similar to the one for the median. First arrange the n data values in ascending order, and then compute the location (or index, i ) of the kth percentile using the formula:

If i is an integer, the ith data value is the kth  percentile. If i is not an integer, take the mean of the two values on either side of i to give the kth percentile.

You’ve already seen the usefulness of the quartiles of a data set (recall the 5-Number Summary and boxplots from Unit 2?). The quartiles are the 25th, 50th, and 75th percentiles:

• Q1 = 25th percentile
• Q2 = 50th percentile = median
• Q3 = 75th percentile

You can find the quartiles using the same index formula above.

Example: Finding the z-Value That Represents a Given Percentile

What z-value represents the 90th percentile?

Solution

The 90th percentile is the z-value for which 90% of the area under the standard normal curve is to the left of z. So, we need to find the value of z that has an area of 0.9000 to its left.

Looking for 0.9000 (or an area extremely close to it) in the interior of the cumulative normal tables, we find 0.8997, which corresponds to a z-value of 1.28. Thus     z ≈ 1.28 represents the 90th percentile.

Using a TI-83/84 Plus calculator, we can find a value of z with a given area to its left. As noted previously, the 90th percentile is the z-value that has an area of 0.9000 to its left. Enter invNorm(0.9000), as shown in the screenshot, and press  ENTER. The answer is z ≈ 1.28.

## 6. Binomial Distribution

There are many real world scenarios that you may encounter where a situation is repeated several times with the same likelihood of success or failure. For example: a basketball player shooting free throws, tossing a coin repeatedly, or looking for errors in a production line. These repeated experiments are called a binomial experiment.

A binomial experiment has four characteristics:

1. The experiment is repeated a fixed number of times called n, these are called observations or trials.
2. Each trial results in only one of two possible outcomes, which we call either “success” or “failure.”
3. The probability of success on a single trial does not change as we repeat the experiment from trial to trial and is called p. The probability of failure in each trial is then (1-p).
4. The n observations or trials are That is, knowing the result of one trial does not change the probabilities we assign to the other observations.

We are interested in x, the number of successes observed in the first n trials, for x=0, 1, 2, … , n. The count of x successes in a binomial experiment has a binomial distribution. This distribution has parameters n and p, where n is the number of trials and p is the probability of success on one trial. We write that x~B(n,p) or x~Bin(n,p), to say that x has such a distribution.

For example a light bulb producer might be interested in how likely it is that he has shipped a defective bulb. He takes 10 light bulbs and then test them all, instead of just testing one light bulb. If each light bulb had a probability of 0:99 of working, what is the probability that all 10 chosen light bulbs work? This is a question that can be answered with the binomial distribution. To see how to calculate the distribution we will use an example

##### Example:

Suppose as an experiment you flip a coin four times and record the results. You are interested in how likely you are to get two heads in four flips, since this would probably mean that the coin is fair.

For us the success is getting a heads, but there are lots of ways for this to happen. This could be happen by getting any of the flips in the following sequence:

{HHTT; HTHT; HTTH; THHT; THTH; TTHH}

We see that there are 6 ways to get exactly 2 heads when flipping a coin 4 times. Thus there are 6 ways to get exactly two heads, and there are 24 = 16 total outcomes in flipping a coin 4 times. Combining these calculations we get that the probability of getting two heads is 6 out of 16 or:

But a more useful way to see this as follows:

It is important to notice that here both our probability calculations are the same for all 6 possibilities. There are two successes and two failures. The more difficult part was finding out the number of ways we could arrange two successes in four trials, or figuring out there were 6 ways to arrange two heads in four flips. However, how to count this is known and it is called a binomial coefficient.

The binomial coefficient is the number of ways to arrange k successes among n observations and is given by the formula:

Where k=0, 1, 2, … ,n.

Usually  is read as n choose k and n! is read as “n factorial”. Factorials are successively multiplying and reducing until 1 is reached, so 6! = 6 ∙  5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720.

For our heads experiment we would have calculated this as:

This result allows us to easily describe the binomial probabilities. In our example with the use of binomial coefficients would give us:

Additionally, this result will allow us to find the distribution of the binomial random variable.

#### Binomial Mean and Standard Deviation

If x represents the count of successes in a binomial experiment with n trials and probability of success p, then what should be the mean? Using reasoning we can picture an experiment where we know how often there is success, p and how many times we repeat the trials n times. We should expect that on average the number of success should be np. For example if a light bulb works with .99 probability and we turn it on 200 times, we expect to have 198 success. So if x has a binomial distribution with mean μ=np and standard deviation

Example (from Moore’s Basic Practice of Statistics)

Typing errors fall in to two categories: “nonword” errors (typing teh for the), which will be caught by a word processor and word errors (using form where the word from goes), which will not be caught by a word processor. Most human proofreaders will catch about 70% of all word errors.  You ask a fellow student to check an essay for you, in which you deliberately placed 10 word errors.

• (a) What is the probability that the student will catch exactly 2 errors?
• (b) What is the probability that the student will catch 2 or fewer errors?
• (c) What is the mean number of missed errors caught?
• (d) What is the standard deviation or the number of errors caught?
• (e) Suppose that the proofreader catches 90% of errors, what is the standard deviation then? What about 99%? What happens to the standard deviation as the probability of success increases toward 1?

Solution

• (a) This is a binomial experiment with n=10. If we define our success to be the student catches the error then p=.7. Since we want exactly two errors we need to find the probability that x=2.

• (b)
• (c) μ = np = 10(.7) = 7  or on average they catch 7 errors. This justifies that small probability in a and b.
• (d)
• (e)
σ decreases toward 0 as p increases toward 1.