Chapter 4, Problem 70CR

Suppose that the distribution of scores on an exam can be described by a normal curve with mean 100. The 16th percentile of this distribution is 80.

1. a. What is the 84th percentile?
2. b. What is the approximate value of the standard deviation of exam scores?
3. c. What z score is associated with an exam score of 90?
4. d. What percentile corresponds to an exam score of 140?
5. e. Do you think there were many scores below 40? Explain.

To determine

Find the 84^th percentile of the distribution.

Answer to Problem 70CR

The 84^th percentile of the distribution is 120.

Explanation of Solution

Calculation:

It is assumed that the exam scores were distributed to normal with mean 100 to its corresponding standard deviation of 80. It is also given that the 16^th percentile of the distribution is 80.

The ${84}^{\text{th}}\left(={100}^{\text{th}}-{16}^{\text{th}}\right)$ percentile of the distribution will have the distance above the mean same as the 16^th percentile have, whose value is below the mean.

The value of the 16^th percentile is 20 units below the mean. In the same way, the value of the 84^th percentile is 20 units above the mean, which will be obtained as given below:

$\begin{array}{c}{84}^{\text{th}}\text{ percentile}=\text{mean}+20\\ =100+20\\ =120\end{array}$

Therefore, the 84^th percentile of the distribution is 120.

To determine

Find the approximate value of the standard deviation of the exam scores.

Answer to Problem 70CR

The approximate value of the standard deviation of the exam scores is 20.

Explanation of Solution

Calculation:

From Part (a), it is observed that the percentage of the exam scores below 80 is 16% and the percentage of the exam scores above 120 is 16%.

The percentage of the exam scores between 80 and 120 is given below:

$\begin{array}{c}\text{percentage}=\text{100}-2\left(\begin{array}{l}\text{percentage of scores above 120 or}\\ \text{percentage of scores below 80}\end{array}\right)\\ =100\%-2\left(16\%\right)\\ =100\%-32\%\\ =68\%\end{array}$

Empirical Rule:

The empirical rule for a normal model states as follows:

• Within 1 standard deviation of mean, 68% of all observations will lie.
• Within 2 standard deviations of mean, 95% of all observations will lie.
• Within 3 standard deviations of mean, 99.7% of all observations will lie.

Based on the empirical rule, 68% of the observation will fall between 1 standard deviation about mean. Therefore, the range of the values that falls within 1 standard deviation is 80 and 120.

The approximate value of the standard deviation is obtained as given below:

$\begin{array}{c}\text{Standard deviation}=\frac{\left\{\begin{array}{l}\text{width of the observation}\\ \text{within 1 standard deviation}\end{array}\right\}}{2}\\ =\frac{120-80}{2}\\ =\frac{40}{2}\\ =20\end{array}$

The above calculation states that the value of standard deviation is 20 units approximately.

To determine

Find the z-score for the exam score of 90.

Answer to Problem 70CR

The z-score for the exam score of 90 is –0.5.

Explanation of Solution

Calculation:

Standardized z-score:

Standardized z-score of an observation is defined as the number of standard deviations the observation is away from the mean. The standardized z-score is also known as the standard score. The standardized z-score is obtained as follows:

$z=\frac{\text{Observation}-\text{mean}}{\text{standard deviation}}$.

The standardized z-score for the exam score of 90 is obtained as given below:

In this case, observation = 90.

Substitute the corresponding value to get the z-score.

$\begin{array}{c}z=\frac{90-100}{20}\\ =\frac{-10}{20}\\ \approx -0.5\end{array}$

Thus, the standardized z-score is –0.5.

The z-score of –0.5 implies that the observation of 90 is 0.5 standard deviations below the mean score of 100.

To determine

Find the percentile that corresponds to an exam score of 140.

Answer to Problem 70CR

The percentile that corresponds to an exam score of 140 is 97.5.

Explanation of Solution

Using empirical rule, the range of the values that falls within 2 standard deviations is given below:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 2 standard deviation }\\ \text{away from the mean}\end{array}\right\}=\text{mean}\pm 2\text{standard devition}\\ =100\pm 2\left(20\right)\\ =100\pm 40\\ =\left(100-40,100+40\right)\\ =\left(60,140\right)\end{array}$

Based on the empirical rule, 95% of the observation will fall between 2 standard deviations about mean.

The score of 140 falls above 2 standard deviations from the mean.

However, the empirical rule states that 5% of observation falls above or below the mean. Thus, the proportion of scores, which falls greater than 140, is $2.5\%\left(=\frac{5\%}{2}\right)$. Therefore, the percentile that corresponds to an exam score of 140 is approximately 95 + 2.5 = 97.5.

To determine

State whether it is appropriate to think that “there were many scores below 40.”.

Answer to Problem 70CR

No, it is not appropriate to think that “there were many scores below 40.”

Explanation of Solution

Calculation:

Using empirical rule, the range of the values that falls within 3 standard deviations is given below:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 3 standard deviation }\\ \text{away from the mean}\end{array}\right\}=\text{mean}\pm 3\text{standard devition}\\ =100\pm 3\left(20\right)\\ =100\pm 60\\ =\left(100-60,100+60\right)\\ =\left(40,160\right)\end{array}$

Based on the empirical rule, 99.7% of the observation will fall between 3 standard deviations about mean.

Thus, the score of 40 falls 3 standard deviations below the mean.

The percentage of the scores that falls below 40 is given below:

$\begin{array}{c}\left\{\begin{array}{l}\text{Percentage of scores }\\ \text{that fall below 40}\end{array}\right\}=\frac{\left(100-99.7\right)\%}{2}\\ =\frac{0.3\%}{2}\\ =0.15\%\end{array}$

Therefore, the percentage of the scores that falls below 40 is only 0.15%.

Thus, it is not appropriate to think that “there were many scores below 40”.