Given a normal population whose mean is 630 and whose standa

Given a normal population whose mean is 630 and whose standard deviation is 47, find each of the following: A. The probability that a random sample of 6 has a mean between 633 and 643.B. The probability that a random sample of 17 has a mean between 633 and 643.C. The probability that a random sample of 26 has a mean between 633 and 643.

Solution

Given a normal population whose mean is 630 and whose standard deviation is 47, find each of the following:

Standard error =sd/sqrt(n) =47/sqrt(6) =19.1877

Z value for 633, z=(633-630)/19.1877 =0.16

Z value for 643, z=(643-630)/19.1877 =0.68

P(633<mean x<643)=P(0.16<z<0.68)

= 0.7517 - 0.5636

=0.1881

Standard error =sd/sqrt(n) =47/sqrt(17) =11.3992

Z value for 633, z=(633-630)/ 11.3992=0.26

Z value for 643, z=(643-630)/ 11.3992=1.14

P(633<mean x<643)=P(0.26<z<1.14)

= 0.8729 - 0.6026

=0.2703

C. The probability that a random sample of 26 has a mean between 633 and 643.

Standard error =sd/sqrt(n) =47/sqrt(26) =9.2175

Z value for 633, z=(633-630)/ 9.2175=0.33

Z value for 643, z=(643-630)/ 9.2175 =1.41

P(633<mean x<643)=P(0.33<z<1.41)

= 0.9207-0.6293

=0.2914