# Normal Applications

CO-6: Apply basic concepts of probability, random variation, and commonly used statistical probability distributions.
Video: Normal Applications (9:41)

## Working with Non-standard Normal Values

LO 6.17: Find probabilities associated with a specified normal distribution.

In a much earlier example, we wondered,

“How likely or unlikely is a male foot length of more than 13 inches?” We were unable to solve the problem, because 13 inches didn’t happen to be one of the values featured in the Standard Deviation Rule.

Subsequently, we learned how to standardize a normal value (tell how many standard deviations below or above the mean it is) and how to use the normal calculator or table to find the probability of falling in an interval a certain number of standard deviations below or above the mean.

By combining these two skills, we will now be able to answer questions like the one above.

To convert between a non-standard normal (X) and the standard normal (Z) use the following equations, as needed: ## EXAMPLE: Male Foot Length

Male foot lengths have a normal distribution, with mean (mu, μ) = 11 inches, and standard deviation (sigma, σ) = 1.5 inches.

(a) What is the probability of a foot length of more than 13 inches? First, we standardize: The probability that we seek, P(X > 13), is the same as the probability that a normal variable takes a value greater than 1.33 standard deviations above its mean, i.e. P(Z > +1.33) This can be solved with the normal calculator or table, after applying the property of symmetry:

P(Z > +1.33) = P(Z < -1.33) = 0.0918.

A male foot length of more than 13 inches is on the long side, but not too unusual: its probability is about 9%.

We can streamline the solution in terms of probability notation and write:

P(X > 13) = P(Z > 1.33) = P(Z < −1.33) = 0.0918

(b) What is the probability of a male foot length between 10 and 12 inches? The standardized values of 10 and 12 are, respectively, Note: The two z-scores in a “between” problem will not always be the same value. You must calculate both or, in this case, you could recognize that both values are the same distance from the mean and hence result in z-scores which are equal but of opposite signs. P(-0.67 < Z < +0.67) = P(Z < +0.67) – P(Z < -0.67) = 0.7486 – 0.2514 = 0.4972.

Or, if you prefer the streamlined notation,

P(10 < X < 12) = P(−0.67 < Z < +0.67) = P( Z < +0.67) − P(Z < −0.67) = 0.7486 − 0.2514 = 0.4972.

By solving the above example, we inadvertently discovered the quartiles of a normal distribution! P(Z < -0.67) = 0.2514 tells us that roughly 25%, or one quarter, of a normal variable’s values are less than 0.67 standard deviations below the mean.

P(Z < +0.67) = 0.7486 tells us that roughly 75%, or three quarters, are less than 0.67 standard deviations above the mean.

And of course the median is equal to the mean, since the distribution is symmetric, the median is 0 standard deviations away from the mean. Be sure to verify these results for yourself using the calculator or table!

Let’s look at another example.

## EXAMPLE: Length of a Human Pregnancy

Length (in days) of a randomly chosen human pregnancy is a normal random variable with mean (mu, μ) = 266 and standard deviation (sigma, σ) = 16.

(a) Find Q1, the median, and Q3. Using the z-scores we found in the previous example we have

Q1 = 266 – 0.67(16) = 255

median = mean = 266

Q3 = 266 + 0.67(16) = 277

Thus, the probability is 1/4 that a pregnancy will last less than 255 days; 1/2 that it will last less than 266 days; 3/4 that it will last less than 277 days. (b) What is the probability that a randomly chosen pregnancy will last less than 246 days?

Since (246 – 266) / 16 = -1.25, we write

P(X < 246) = P(Z < −1.25) = 0.1056

(c) What is the probability that a randomly chosen pregnancy will last longer than 240 days?

Since (240 – 266) / 16 = -1.63, we write

P(X > 240) = P(Z > −1.63) = P(Z < +1.63) = 0.9484

Since the mean is 266 and the standard deviation is 16, most pregnancies last longer than 240 days.

(d) What is the probability that a randomly chosen pregnancy will last longer than 500 days?

Method 1:

Common sense tells us that this would be impossible.

Method 2:

The standardized value of 500 is (500 – 266) / 16 = +14.625.

P(X > 500) = P(Z > 14.625) = 0.

(e) Suppose a pregnant woman’s husband has scheduled his business trips so that he will be in town between the 235th and 295th days. What is the probability that the birth will take place during that time?

The standardized values are (235 – 266) / 16) = -1.94 and (295 – 266) / 16 = +1.81.

P(235 < X < 295) = P(−1.94 < Z < +1.81) = P(Z < +1.81) − P(Z < −1.94) = 0.9649 − 0.0262 = 0.9387.

There is close to a 94% chance that the husband will be in town for the birth.

Be sure to verify these results for yourself using the calculator or table!

The purpose of the next activity is to give you guided practice at solving word problems that involve normal random variables. In particular, we’ll solve problems like the examples you just went over, in which you are asked to find the probability that a normal random variable falls within a certain interval.

The previous examples most followed the same general form: given values of a normal random variable, you were asked to find an associated probability. The two basic steps in the solution process were to

• Standardize to Z;
• Find associated probabilities using the standard normal calculator or table.

## Finding Normal Scores

LO 6.18: Given a probability, find scores associated with a specified normal distribution.

The next example will be a different type of problem: given a certain probability, you will be asked to find the associated value of the normal random variable. The solution process will go more or less in reverse order from what it was in the previous examples.

## EXAMPLE: Foot Length

Again, foot length of a randomly chosen adult male is a normal random variable with a mean of 11 and standard deviation of 1.5.

(a) The probability is 0.04 that a randomly chosen adult male foot length will be less than how many inches? According to the normal calculator or table, a probability of 0.04 below (actually 0.0401) is associated with z = -1.75. In other words, the probability is 0.04 that a normal variable takes a value lower than 1.75 standard deviations below its mean.

For adult male foot lengths, this would be 11 – 1.75(1.5) = 8.375. The probability is 0.04 that an adult male foot length would be less than 8.375 inches. (b) The probability is 0.10 that an adult male foot will be longer than how many inches? Caution is needed here because of the word “longer.”

Once again, we must remind ourselves that the calculator and table only show the probability of a normal variable taking a value lower than a certain number of standard deviations below or above its mean. Adjustments must be made for problems that involve probabilities besides “lower than” or “less than.” As usual, we have a choice of invoking either symmetry or the fact that the total area under the normal curve is 1. Students should examine both methods and decide which they prefer to use for their own purposes.

Method 1:

According to the calculator or table, a probability of 0.10 below is associated with a z value of -1.28. By symmetry, it follows that a probability of 0.10 above has z = +1.28.

We seek the foot length that is 1.28 standard deviations above its mean: 11 + 1.28(1.5) = 12.92, or just under 13 inches.  Method 2: If the probability is 0.10 that a foot will be longer than the value we seek, then the probability is 0.90 that a foot will be shorter than that same value, since the probabilities must sum to 1.

According to the calculator or table, a probability of 0.90 below is associated with a z value of +1.28. Again, we seek the foot length that is 1.28 standard deviations above its mean, or 12.92 inches. Comment:

• Part (a) in the above example could have been re-phrased as: “0.04 is the proportion of all adult male foot lengths that are below what value?”, which takes the perspective of thinking about the probability as a proportion of occurrences in the long-run. As originally stated, it focuses on the chance of a randomly chosen individual having a normal value in a given interval.

## EXAMPLE: Money Spent for Lunch

A study reported that the amount of money spent each week for lunch by a worker in a particular city is a normal random variable with a mean of \$35 and a standard deviation of \$5.

(a) The probability is 0.97 that a worker will spend less than how much money in a week on lunch?

The z associated with a probability of 0.9700 below is +1.88. The amount that is 1.88 standard deviations above the mean is 35 + 1.88(5) = 44.4, or \$44.40.

(b) There is a 30% chance of spending more than how much for lunches in a week?

The z associated with a probability of 0.30 above is +0.52. The amount is 35 + 0.52(5) = 37.6, or \$37.60.

Comment:

• Another way of expressing Example (part a.) above would be to ask, “What is the 97th percentile for the amount (X) spent by workers in a week for their lunch?” Many normal variables, such as heights, weights, or exam scores, are often expressed in terms of percentiles.

## EXAMPLE:

The height X (in inches) of a randomly chosen woman is a normal random variable with a mean of 65 and a standard deviation of 2.5.

What is the height of a woman who is in the 80th percentile?

A probability of 0.7995 in the table corresponds to z = +0.84. Her height is 65 + 0.84(2.5) = 67.1 inches.

By now we have had practice in solving normal probability problems in both directions: those where a normal value is given and we are asked to report a probability and those where a probability is given and we are asked to report a normal value. Strategies for solving such problems are outlined below:

• Given a normal value x, solve for probability:
• Standardize: calculate • If you are using the online calculator: Type the z-score for which you wish to find the area to the left and hit “compute.”
• If you are using the table: Locate z in the margins of the normal table (ones and tenths for the row, hundredths for the column). Find the corresponding probability (given to four decimal places) of a normal random variable taking a value below z inside the table.
• (Adjust if the problem involves something other than a “less-than” probability, by invoking either symmetry or the fact that the total area under the normal curve is 1.)
• Given a probability, solve for normal value x:
• (Adjust if the problem involves something other than a “less-than” probability, by invoking either symmetry or the fact that the total area under the normal curve is 1.)
• Locate the probability (given to four decimal places) inside the normal table. Using the table, find the corresponding z value in the margins (row for ones and tenths, column for hundredths). Using the calculator, provide the area to left of the z-score you wish to find and hit “compute.”
• “Unstandardize”: calculate This next activity is a continuation of the previous one, and will give you guided practice in solving word problems involving the normal distribution. In particular, we’ll solve problems like the ones you just solved, in which you are given a probability and you are asked to find the normal value associated with it.

Learn by Doing: Find Normal Scores

## Normal Approximation for Binomial

The normal distribution can be used as a reasonable approximation to other distributions under certain circumstances.  Here we will illustrate this approximation for the binomial distribution.

We will not do any calculations here as we simply wish to illustrate the concept.  In the next section on sampling distributions, we will look at another measure related to the binomial distribution, the sample proportion, and at that time we will discuss the underlying normal distribution.

Consider the binomial probability distribution displayed below for n = 20 and p = 0.5. Now we overlay a normal distribution with the same mean and standard deviation. And in the final image, we can see the regions for the exact and approximate probabilities shaded. Unfortunately, the approximated probability, 0.1867, is quite a bit different from the actual probability, 0.2517. However, this example constitutes something of a “worst-case scenario” according to the usual criteria for use of a normal approximation.

## Rule of Thumb

Probabilities for a binomial random variable X with n and p may be approximated by those for a normal random variable having the same mean and standard deviation as long as the sample size n is large enough relative to the proportions of successes and failures, p and 1 – p. Our Rule of Thumb will be to require that

np ≥ 10 and n(1 − p) ≥ 10

## Continuity Correction

It is possible to improve the normal approximation to the binomial by adjusting for the discrepancy that arises when we make the shift from the areas of histogram rectangles to the area under a smooth curve. For example, if we want to find the binomial probability that X is less than or equal to 8, we are including the area of the entire rectangle over 8, which actually extends to 8.5. Our normal approximation only included the area up to 8.