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-Dominant- [34]
3 years ago
7

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
Mathematics
1 answer:
patriot [66]3 years ago
6 0

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

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