Chapter 11.10, Problem 31PPSE

Solution

To explain: Whether Jake and Elena scored top 10% of their group in a standardized test.

Jake score is not in the top 10% of his group but Elena scored in the top 10% of her group.

Given:

The score of Jake and Elena in a standardized test in different classes $=87$

The mean and standard deviation of Jake’s group $\text{(mean, standard deviation)}=(80,6)$

The mean and standard deviation of Elena’s group $\text{(mean, standard deviation)}=(76,4)$

Concept used:

The probabilities of data less than a particular score in a normal distribution is calculated using Z scores. Z score is calculated using the formula $Z=\frac{X-\mu }{\sigma }$

Calculation:

The Z score for Jake’s test score is calculated as shown below

  $\begin{array}{l}Z=\frac{X-\mu }{\sigma }\\ Z=\frac{87-80}{6}=1.17\end{array}$

The Z score for Jake’s test score with respect to the mean and standard deviation of his group is 1.17 (rounded to two decimal places).

The Z score for Elena’s test score is calculated as shown below

  $\begin{array}{l}Z=\frac{X-\mu }{\sigma }\\ Z=\frac{87-76}{4}=2.75\end{array}$

The Z score for Elena’s test score with respect to the mean and standard deviation of her group is 2.75 (rounded to two decimal places).

With respect to the Z table the probability less than Z score 1.17 is $P(Z<1.17)=0.8790$ .

The probability less than Z score 2.75 is $P(Z<2.75)=0.9970$ .

Interpretation:

The percentage of students who scores less than Jake in his group is 87.9%. The percentage of students who scores less than Elena in her group is 99.7%. Thus Jake is not in the top 10% of students in his group whereas Elena is in the top 10% of students in her group.