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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

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JeanaShupp

Answer: 0.0775

Step-by-step explanation:

Given : Mean : [tex]\mu = 25[/tex]

Standard deviation : [tex]\sigma =15[/tex]

Sample size : [tex]n=36[/tex]

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

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x= 28.2 hours

[tex]z=\dfrac{28.2-25}{\dfrac{15}{\sqrt{36}}}=1.28[/tex]

For x= 30 hours

[tex]z=\dfrac{30-25}{\dfrac{15}{\sqrt{36}}}=2[/tex]

The P- value = [tex]P(1.28<z<2)[/tex]

[tex]=P(z<2)-P(z<1.28)= 0.9772498-0.8997274=0.0775224\approx0.0775[/tex]

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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