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

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=(x-\mu)/((\sigma)/(√(n)))`

For x= 28.2 hours

`z=(28.2-25)/((15)/(√(36)))=1.28`

For x= 30 hours

`z=(30-25)/((15)/(√(36)))=2`

The P- value =
`P(1.28<z<2)`

`=P(z<2)-P(z<1.28)= 0.9772498-0.8997274=0.0775224\approx0.0775`

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