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A population has a mean of 300 and a standard deviation of 90. Suppose a sample of size 100 is selected and  is used to estimate . Use z-table.

1. What is the probability that the sample mean will be within +/- 5 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)
   
   
2. What is the probability that the sample mean will be within +/- 13 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answer 1

Given

Sample size=n=100

Population standard deviation==90

Sample standard deviation (s) is given by

i)

Let X denotes that sample mean. We are required to find that the sample mean is between -5 (300-5=295) and +5 (300+5) of population mean i.e.

ii)

We are required to find that the sample mean is between -13 (300-13=287) and +13 (300+13) of population mean i.e.