Chapter 3, Problem 3.49E

a.

To determine

To draw: A histogram with a superimposed normal curve.

Answer to Problem 3.49E

The obtained histogram with a superimposed normal curve is,

Explanation of Solution

Given info:

The data of SATMATH are given.

Calculation:

Histogram:

Software Procedure:

Step by step procedure to obtain Histogram is given as:

• Choose Graph > Histogram.
• Choose With Fit, and then click OK.
• In Graph variables, enter the corresponding data column SATMATH.
• Click OK.

The output using Minitab software is given as:

b.

To determine

To find: The mean, median, standard deviations and quartiles for the given data.

To compare: The mean and the median.

To compare: The quartiles.

Answer to Problem 3.49E

The mean, median, standard deviation, 1^st quartile, 2^nd quartile are 543.2, 540, 61.69, 500 and 580, respectively.

The distribution is symmetric.

Explanation of Solution

Calculation:

Descriptive statistics:

Software Procedure:

Step by step procedure to obtain the inter quartile range, Q1 and Q3 using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns SAT.
• In statistics choose Mean, Median, Standard deviation, 1^st quartile, 3^rd quartile..
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output,

The mean is 543.2.

The standard deviation is 61.69.

The 1^st quartile is 500.

The median is 540.

The 3^rd quartile is 580.

Comparison between mean and median:

If mean is equal to the median then it is said to be symmetric.

Here, the mean is close to the median. Thus, the distribution is symmetric.

Distance of the first quartile from the median:

The formula to find the distance of the first quartile from the median is,

$\begin{array}{c}\text{Median}-\text{First Quartile = }540-500\\ =40\end{array}$

Thus, the distance of the first quartile from the median is 40.

Distance of the first quartile from the median:

The formula to find the distance of the third quartile from the median is,

$\begin{array}{c}\text{Third Quartile}-\text{Median = }580-540\\ =40\end{array}$

Thus, the distance of the third quartile from the median is 40.

Here, the distance of the first quartile from the median is approximately equal to distance of the third quartile from the median. That is, $\left(40\right)$. Thus, the distribution is symmetric.

Hence, the distribution of the rainwater specimens is quite symmetric.

c.

To determine

To find: The proportion of GSU freshmen scored above the mean 514 for all college-bound seniors.

Answer to Problem 3.49E

The proportion of GSU freshmen scored above the mean 514 for all college-bound seniors is 0.6808.

Explanation of Solution

The formula to find the standardized score is,

$z=\frac{x-\mu }{\sigma }$

In 2013, the mean score on the mathematics portion of the SAT for all college-bound seniors was 514.

The distribution of first 20 SAT mathematics scores follow normal distribution with mean 543.2 and standard deviation 61.69.

Consider the random variable X, denotes those students of first 20 SAT mathematics scores.

Thus,

$\begin{array}{c}P\left(X>514\right)=P\left(Z>\frac{514-543.2}{61.69}\right)\\ =P\left(Z>-0.47\right)\end{array}$

It is known that, $P\left(0\le Z<\infty \right)=0.5$ and $P\left(-\infty <Z\le 0\right)=0.5$.

Thus,

$\begin{array}{c}P\left(Z>-0.47\right)=1-P\left(Z\le -0.47\right)\\ =1-P\left(-\infty <Z\le -0.47\right)\end{array}$

Cumulative probability value:

Software Procedure:

Step by step procedure to obtain the graph using the MINITAB software:

• Choose Calc > Probability distribution> Normal.
• Choose Cumulative Probability.
• In mean, enter 0.
• In standard deviation, enter 1.
• Choose Input Constant.
• In Input Constant, enter -0.47.
• Click OK.

Output using the MINITAB software is given below:

Thus,

$\begin{array}{c}P\left(X>514\right)=1-P\left(Z\le -0.47\right)\\ =1-0.319178\\ =0.6808\end{array}$

Therefore, the proportion of GSU freshmen scored above the mean for all college-bound seniors is 0.6808.

d.

To determine

To find: The exact proportion of GSU freshmen who scored above the mean for all college-bound seniors.

Answer to Problem 3.49E

The exact proportion of GSU freshmen who scored above the mean for all college-bound seniors is 0.6840.

Explanation of Solution

It is found from the data that 1,859 students, out of a total of 2,718 students have scored above the mean score 514 for all college-bound seniors.

Thus, the exact proportion of GSU freshmen who scored above the mean for all college-bound seniors is:

$\begin{array}{c}\frac{1,859}{2,718}=0.68395\\ \approx \mathrm{0.6840.}\end{array}$

From part (c), the proportion of GSU freshmen scored above the mean 514 for all college-bound seniors, for normal distribution is 0.6808 or 68.08%.

Here, the exact proportion of GSU freshmen who scored above the mean for all college-bound seniors is 0.6840 or 68.40%.

It can be said that $68.08\%\approx 68.10\%$, which is quite close to 68.40%.

Thus, the exact percentage obtained from part (d) and the percentage obtained from part (c) are very close.

Thus, despite any discrepancies, the distribution can be considered to be “close enough to Normal” distribution.