AP Statistics 2.11: Normal Distribution

Use normal models to describe shape, calculate interval probabilities, and find percentiles or measurement bounds from areas.

Covered syllabus points

2.11.A—Describe a normal distribution 2.11.B—Calculate the mean and standard deviation for a normal distribution 2.11.C—Calculate percentages from a normal distribution using the empirical rule 2.11.D—Calculate the probability that a particular value lies within a given interval of a normal distribution 2.11.E—Calculate the associated intervals and areas of a normal distribution 2.11.F—Compare measures of relative position for distributions

A professional sports team evaluates potential players for a certain position based on two main characteristics, speed and strength.

Speed is measured by the time required to run a distance of 40 yards, with smaller times indicating more desirable (faster) speeds. From previous speed data for all players in this position, the times to run 40 yards have a mean of 4.60 seconds and a standard deviation of 0.15 seconds, with a minimum time of 4.40 seconds, as shown in the table below.

Table for Question 2.11 The Normal Distribution question 1 — AP Statistics

Based on the relationship between the mean, standard deviation, and minimum time, is it reasonable to believe that the distribution of 40-yard running times is approximately normal? Explain.

Part (a):
No, it is not reasonable to believe that the distribution of 40-yard running times is approximately normal, because the minimum time is only 1.33 standard deviations below the mean $\left(z=\frac{4.4-4.6}{0.15} \approx-1.33\right)$ . In a normal distribution, approximately 9.2 percent of the z-scores are below -1.33. However, there are no running times less than 4.4 seconds, which indicates that there are no running times with a z-score less than -1.33. Therefore, the distribution of 40 -yard running times is not approximately normal.

[Maximum number: 4]

Each full carton of Grade A eggs consists of 1 randomly selected empty cardboard container and 12 randomly selected eggs. The weights of such full cartons are approximately normally distributed with a mean of 840 grams and a standard deviation of 7.9 grams.

(a)

What is the probability that a randomly selected full carton of Grade A eggs will weigh more than 850 grams?

Part (a):
Let W denote the weight of a randomly selected full carton of eggs. W has a normal distribution with mean 840 grams and standard deviation 7.9 grams.

The z-score for a weight of 850 grams is $z=\frac{850-840}{7.9} \approx 1.27$ .
The standard normal probability table reveals that

P(W>850)=P(Z>1.27) \approx 1-0.8980=0.1020 .

(b)

The weights of the empty cardboard containers have a mean of 20 grams and a standard deviation of 1.7 grams. It is reasonable to assume independence between the weights of the empty cardboard containers and the weights of the eggs. It is also reasonable to assume independence among the weights of the 12 eggs that are randomly selected for a full carton.
Let the random variable X be the weight of a single randomly selected Grade A egg.

(i)

What is the mean of X ?

Let W represent the weight of a randomly selected full carton of eggs, P the weight of the packaging, and $X_{i}$ the weight of the i th egg, for $i=1,2 \ldots, 12$ .
Note that $W=P+X_{1}+X_{2}+\ldots+X_{12}$ .
Properties of expected values establish that $\mathrm{E}(W)=\mathrm{E}(P)+\mathrm{E}\left(X_{1}\right)+\ldots+\mathrm{E}\left(X_{12}\right)$ .
Because all 12 eggs have the same mean weight, this becomes $\mathrm{E}(W)=\mathrm{E}(P)+12 \times \mathrm{E}\left(X_{i}\right)$ .
We were told that E(W)=840 and E(P)=20, so we can solve

840=20+12 \times \mathrm{E}\left(X_{i}\right) \text { to find } \mathrm{E}\left(X_{i}\right)=\frac{840-20}{12} \approx 68.33 \text { grams. }

(ii)

What is the standard deviation of X ?

Because of independence, properties of variance establish that
Var(W)=Var(P)+Var(X_1)+Var(X_2)++Var(X_12).
Because all 12 eggs have the same variance of their weights, this becomes
Var(W)=Var(P)+12 x Var(X_i).
We were told that $\operatorname{SD}(W)=7.9$ and $\operatorname{SD}(P)=1.7$ . Therefore, $\operatorname{Var}(W)=(7.9)^{2}=62.41$ and $\operatorname{Var}(P)=(1.7)^{2}=2.89$ .
We can solve $62.41=2.89+12 \times \operatorname{Var}\left(X_{i}\right)$ to find $\operatorname{Var}\left(X_{i}\right)=\frac{62.41-2.89}{12}=4.96$ . Thus, $\operatorname{SD}\left(X_{i}\right)=\sqrt{(4.96)} \approx 2.23$ grams.

[Maximum number: 1]

The distribution of the sale price of a certain car model is approximately normal with a mean of 65,500 and a standard deviation of $3,100. Based on the distribution, which of the following is an appropriate conclusion?

Approximately 95\% of the cars of this model have a sale price of less than $66,700.

Approximately 99\% of the cars of this model have a sale price between $57,500 and $73,500.

The interquartile range of the car model's sale price is approximately $6,200.

The maximum of the car model's sale price is $69,800.

Approximately 84\% of the cars of this model have a sale price greater than $57,400. A scam call is a call made to deceive the person receiving the call. A report claims that 45 percent of all calls received on mobile phones can be classified as scam calls. A mobile phone service believes that the percent of scam calls is different than the 45 percent claimed in the report. Let P represent the population of mobile phone calls that can be classified as scam calls. A random sample of calls received on mobile phone devices was selected in order to construct a 95 percent confidence interval for estimating P.

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Covered syllabus points

2.11 The Normal Distribution question 1

2.11 The Normal Distribution question 2

Question (a)

Question (b)

Question (i)

Question (ii)

2.11 The Normal Distribution question 3