Based on past data the sample mean of the credit card purcha

Based on past data, the sample mean of the credit card purchases at a large department store is $35. Assuming sample size is 25 and the population standard deviation is 10.

a)    What % of samples are likely to have between 20 and 30?   _________

b)    Between what two values 90% of sample means fall?                    ____________

c)     Below what value 99% of sample means fall?

d)    Above what value only 1% of sample means fall??

Within what symmetrical limits of the population percentage will 95% of the sample percentages fall? ___________

Solution

Based on past data, the sample mean of the credit card purchases at a large department store is $35. Assuming sample size is 25 and the population standard deviation is 10.

a)    What % of samples are likely to have between 20 and 30?   24.17%

z value for 20, z=(20-35)/10 = -1.5

z value for 30, z=(30-35)/10 = -0.5

P( 20<x<30) = P( -1.5<z<-0.5) = P( z < -0.5) – P( Z< -1.5)

= 0.3085 - 0.0668 =0.2417

b)    Between what two values 90% of sample means fall?                  (31.71, 38.29)

Standard error = sd/sqrt(n) =10/sqrt(25) =2

Z value for 90% =1.645

Lower value =35-1.645*2 =31.71

upper value =35-1.645*2 =38.29

c)     Below what value 99% of sample means fall?

Z value for 99% =2.326

The required value = 35+2.326*2 = 39.652

d)    Above what value only 1% of sample means fall??

The required value = 35+2.326*2 = 39.652

Within what symmetrical limits of the population percentage will 95% of the sample percentages fall?

Z value for 95% =1.96

Lower value =35-1.96*10 = 15.4

Upper value =35+1.96*10 =54.6