Question

A survey among freshman at a certain university revealed that the number of hours spent studying before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probabiliy that the average time spent stydying for the sampe was between 28.2 and 30 hours

Answer 1

Answer: 0.0775

Step-by-step explanation:

Given : Mean : $\mu = 25$

Standard deviation : $\sigma =15$

Sample size : $n=36$

Since its normal distribution , then the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 28.2 hours

$z=\dfrac{28.2-25}{\dfrac{15}{\sqrt{36}}}=1.28$

For x= 30 hours

$z=\dfrac{30-25}{\dfrac{15}{\sqrt{36}}}=2$

The P- value = $P(1.28$

$=P(z$

Hence, the probabiliy that the average time spent stydying for the sampe was between 28.2 and 30 hours = 0.0775