Chapter 7, Problem 7.59SE

a.

To determine

To make: The stem plot for IQ test scores of 31-seventh grade girls.

To check: Whether there are major departures from normality or not.

Answer to Problem 7.59SE

The stem plot is,

No, there are no major departures from normality.

Explanation of Solution

Given info:

The data shows the IQ test scores of 31 seventh-grade girls in a Midwest school district.

Calculation:

Stemplot:

Software procedure:

Step-by-step software procedure to draw stemplot using MINITAB software is as follows:

• Select Graph > Stem and leaf.
• Select the column of IQ Test Scores in Graph variables.
• Select OK.

The $1.5\times IQR$ rule for outliers:

A observation is a suspected outlier, if it is more than $Q_3+\left(1.5\times IQR\right)$ or less than $Q_1-\left(1.5\times IQR\right)$.

Interquartile range (IQR):

The difference between the first quartile and third quartile is considered as interquartile range.

That is,

$IQR=Q_3-Q_1$

Software procedure:

Step-by-step software procedure for the first quartile and third quartile in MINITAB software is as follows:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns IQ Test Scores.
• Choose option statistics, and select first quartile, third quartile.
• Click OK.

Output using MINITAB software is as follows:

From Minitab output, the first quartile is 98.00 and third quartile is 114.00.

Substitute 114.00 for $Q_3$ and 98.00 for $Q_1$ in interquartile range

The interquartile range is,

$\begin{array}{c}IQR=114.00-98.00\\ =16\end{array}$

Substitute IQR in the $1.5\times IQR$ rule

$\begin{array}{c}{Q}_3+\left(1.5\times IQR\right)=114.00+\left(1.5\times 16\right)\\ =114.00+24\\ =138\end{array}$

$\begin{array}{c}{Q}_1-\left(1.5\times IQR\right)=98.00-\left(1.5\times 16\right)\\ =98.00-24\\ =74\end{array}$

The $1.5\times IQR$ rule suspects one outlier in the IQ test scores of 31 seventh-grade girls because there is a value which is not in between limits $Q_1-\left(1.5\times IQR\right)=74$ and $Q_3+\left(1.5\times IQR\right)=138$.

Justification:

The stemplot showing that there are two low IQ’s. They are 72 and 74but for the small data, the IQ test scores of 31 seventh-grade girls is approximately normal.

b.

To determine

To obtain: The mean $\overline{x}$ and standard deviation s and the proportions of the scores are within 1 standard deviation and within 2 standard deviations of the mean.

To check: Whether the proportions show an exactly normal distribution.

Answer to Problem 7.59SE

The mean $\overline{x}$ is 105.84 and standard deviation s is 14.27.

The proportions of the scores are within 1 standard deviation of the mean, which is 0.742.

The proportions of the scores are within 2 standard deviation of the mean, which is 0.935.

If the distribution is exactly normal, then its proportion should be about 68% of the observations fall within $\sigma$ of the mean $\mu$ and about 95% of the observations fall should be within $2\sigma$ of the mean $\mu$.

Explanation of Solution

68-95-99.7 Rule:

If the proportions be in an exactly normal distribution then it is,

About 68% of the observations fall within $\sigma$ of the mean $\mu$.

About 95% of the observations fall within $2\sigma$ of the mean $\mu$.

About 99.7% of the observations fall within $3\sigma$ of the mean $\mu$.

Calculation:

For Mean and Standard deviation:

Software procedure:

Step-by-step software procedure for mean and standard deviation in MINITAB software is as follows:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns IQ Test Scores.
• Choose option statistics, and select mean, standard deviation.
• Click OK.

Output using MINITAB software is as follows:

From Minitab output, the mean is 105.84 and standard deviation is 14.27.

The scores are within1 standard deviation and within 2 standard deviations of the mean that are obtained by finding $x\pm 1s$ and $x\pm 2s$.

Substitute 105.84 for x and 14.27 for s,

$\begin{array}{c}x+1s=105.84+1\times 14.27\\ x+1s=120.11\end{array}$

$\begin{array}{c}x-1s=105.84-1\times 14.27\\ x-1s=91.57\end{array}$

$\begin{array}{c}x+2s=105.84+2\times 14.27\\ x+2s=134.38\end{array}$

$\begin{array}{c}x-2s=105.84-2\times 14.27\\ x-2s=77.3\end{array}$

For proportion of scores within 1 standard deviation:

The IQ test scores in between $\left[91.57,120.11\right]$ is 23 and the total number IQ test scores is 31.

The formula to find the proportion of scores within 1 standard deviation is,

$\mathrm{Proportion}\text{ of IQ scores = }\frac{\text{Number of IQ test scores in between }\left[91.57,120.11\right]}{\text{Total number of IQ scores}}$

Substitute 23 for ‘Number of IQ test scores in between $\left[91.57,120.11\right]$ and 31 for ‘Total number of IQ scores’.

$\begin{array}{c}\mathrm{Proportion}\text{ of IQ scores =}\frac{\text{23}}{31}\\ =0.7419\\ \approx 0.742\end{array}$

Thus, the proportion of scores within 1 standard deviation is 0.742.

For proportion of scores within 2 standard deviations:

The IQ test scores in between $\left[77.3,134.38\right]$ is 29 and the total number IQ test scores is 31.

The formula to find the proportion of scores within 2 standard deviations is,

$\mathrm{Proportion}\text{ of IQ scores = }\frac{\text{Number of IQ test scores in between }\left[77.3,134.38\right]}{\text{Total number of IQ scores}}$

Substitute 29 for ‘Number of IQ test scores in between $\left[77.3,134.38\right]$ and 31 for ‘Total number of IQ scores’.

$\begin{array}{c}\mathrm{Proportion}\text{ of IQ scores =}\frac{\text{29}}{31}\\ =0.935\end{array}$

Thus, the proportion of scores within 2 standard deviations is 0.935.

Justification:

If the distribution is exactly normal, then it should satisfy about 68% of the observations fall within $\sigma$ of the mean $\mu$ and about 95% of the observations fall within $2\sigma$ of the mean $\mu$.

Here, the proportion of scores within 1 standard deviation is 74.2% and the proportion of scores within 2 standard deviations is 93.5%. Thus, the proportions of scores are reasonably close to the proportion of exactly normal distribution.