I dont have calculator to do this, so plz help me In the Department of Education

Question

I dont have calculator to do this, so plz help me

In the Department of Education at UR University, student records suggest that the population of students spends an average of 5.5 hours per week playing organized sports. The population's standard deviation is 2.2 hours per week. Based on a sample of 121 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.

A) Compute the standard error of the sample mean. (Round your answer to 1 decimal place.)

B) What is the chance HLI will find a sample mean between 5 and 6 hours? (Round z value to 2 decimal places. Round your answer to 4 decimal places.)

C) Calculate the probability that the sample mean will be between 5.3 and 5.7 hours. (Round z value to 2 decimal places. Round your answer to 4 decimal places.)

Explanation / Answer

In the department of education at UR university, student records suggest that the population of students spends an average of 5.5 hrs per week playing organized sports. The population's standard deviation is 2.2 hrs per week.
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Based on a sample of 121 students, healthy lifestyles Inc.
(HLI) would like to apply the central limit theorem to make various estimates.
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Compute the standard error of the sample mean?
Ans: 2.2/sqrt(121) = 0.2
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What is the chance HLI will find a sample mean between 5 and 6 hrs?
t(5) = (5-5.5)/0.2 = -2.5
t(6) = (6-5.5)/0.2 = +2.5
P(5< x-bar <6) = P(-2.5< t < 2.5 when df=120) = 0.9862
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Calculate the probability that the sample mean will be between 5.3 and 5.7 hrs

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z(5.3) = (5.3-5.5)/[2.2/sqrt(121)] = -1
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z(5.7) = (5.7-5.5)/(2.2/sqrt(121)] = +1
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P(5.3< x-bar <5.7) = P(-1< z < 1) = 0.6827
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