2probability and stats 1st question: suppose a sample of 85 is taken from a population with a mean that is claimed to be 38 and a standard deviation of 9. if this is the actual mean of the population, which of these values would be within the 95% confidence interval for the mean of the sample? answer choices: a) 28.2 b) 32.2 c) 36.2 d) 40.2 2nd question: suppose a sample of 102 is taken from a population with a mean that is claimed to be 76 and a standard deviation of 20. if this is the actual mean of the population, which of these values would be outside the 95% confidence interval for the mean of the sample? answer choices: a) 71.8 b) 74.8 c) 76.8 d) 79.8

 1) To find the confidence interval 
the sample mean x = 38     σ = 9; n = 85;
The confidence level is 95% (CL = 0.95) CL = 0.95
so α = 1 – CL = 0.05   
α/2 = 0.025   Z(α/2) = z0.025  
The area to the right of Z0.025 is 0.025 and the area to the left of Z0.025 is 1 – 0.025 = 0.975
Z(α/2) = z0.025 = 1.645 This can be found using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645)*(9)/(85^0.5)=1.6058 x - EBM = 38 – 1.6058 = 36.3941  x + EBM = 38 + 1.6058 = 39.6058
The 95% confidence interval is (36.3941, 39.6058).
The answer is the letter D
The value of 40.2 is within the 95% confidence interval for the mean of the sample

2) To find the confidence interval  
the sample mean x = 76     σ = 20; n = 102; 
The confidence level is 95% (CL = 0.95) CL = 0.95 
so α = 1 – CL = 0.05   
α/2 = 0.025   Z(α/2) = z0.025 
The area to the right of Z0.025 is 0.025 and the area to the left of Z0.025 is 1 – 0.025 = 0.975 
Z(α/2) = z0.025 = 1.645 This can be found using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645)*(20)/(102^0.5)=3.2575 x - EBM = 76 – 3.2575 = 72.7424   x + EBM = 76 + 3.2575 = 79.2575  
The 95% confidence interval is (72.7424 ,79.2575).
The answer is the letter A and the letter D
The value of 71.8 and 79.8  are  outside the 95% confidence interval for the mean of the sample

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