MATH SOLVE

4 months ago

Q:
# Percentage grade averages were taken across all disciplines at a particular university, and the mean average was found to be 83.6 and the standard deviation was 8.7. If 10 classes were selected at random, find the probability that the class average is greater than 90.A. 0.0100B. 0.5247C. 0.1023D. 0.0002

Accepted Solution

A:

Correct Ans:

Option A. 0.0100

Solution:

We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

[tex]z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ z-score =2.326 [/tex]

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

Option A. 0.0100

Solution:

We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

[tex]z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ z-score =2.326 [/tex]

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.