A population has a mean of 300 and a standard deviation of 4

A population has a mean of 300 and a standard deviation of 40. Suppose a sample of size 125 is selected and (x-bar) is used to estimate (mu).

A. What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)?

B. What is the probability that the sample mean will be within +/- 10 of the population mean (to 4 decimals)?

Solution

a)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 300 - 8 =   292      
x2 = upper bound = 300 + 8 =    308      
u = mean =    300      
n = sample size =    125      
s = standard deviation =    40      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.236067977      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.236067977      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.012673659      
P(z < z2) =    0.987326341      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.974652681   [ANSWER]

***********************

b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    290      
x2 = upper bound =    310      
u = mean =    300      
n = sample size =    125      
s = standard deviation =    40      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.795084972      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.795084972      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.002594304      
P(z < z2) =    0.997405696      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.994811392   [ANSWER]