It is known that the certain pull-strength measurement is normally distributed w

Question

It is known that the certain pull-strength measurement is normally
distributed with mean =10 lb. and the standard deviation = 1. 2 lb.

a. What is the probability that a randomly selected measurement is
greater than 10. 525 lb.?

b. What is the probability that the sample mean found from a random
sample of size 10 is greater than 10. 525 lb.?

c. If the population standard deviation is unknown, and s = 1. 2 lb. is the
standard deviation found from the sample, what is the probability that
the sample mean found from a random sample of size 10 is greater
than 10. 525 lb.?

d. If the population standard deviation is unknown, and s = 1. 2 lb. is the
standard deviation found from the sample, what is the probability that
the sample mean found from a random sample of size 50 is greater
than 10. 525 lb.?

Explanation / Answer

a)probability that a randomly selected measurement is
greater than 10. 525 lb =P(X>10.525)=1-P(X<10.525)=1-P(Z<(10.525-10)/1.2)=1-P(Z<0.4375)=1-0.6691=0.3309

b)

for n=10 ; std error of mean =std deviaiton/(n)1/2 =1.2/(10)1/2 =0.3795

hence probability that the sample mean found from a random
sample of size 10 is greater than 10. 525 lb

=P(X>10.525)=1-P(X<10.525)=1-P(Z<(10.525-10)/0.3795)=1-P(Z<0.1.3835)=1-0.9167 =0.0833

c) for unknown population std deviation; we will use t distribution

probability that the sample mean found from a random
sample of size 10 is greater than 10. 525 lb

=P(X>10.525)=1-P(X<10.525)=1-P(t<(10.525-10)/0.3795)=1-P(t<0.1.3835)=1-0.90 =0.10

d)

std error of mean =std deviaiton/(n)1/2 =1.2/(50)1/2 =0.1697

probability that the sample mean found from a random
sample of size 50 is greater than 10. 525 lb

=P(X>10.525)=1-P(X<10.525)=1-P(t<(10.525-10)/0.1697)=1-P(t<0.1.3835)=1-0.9936 =0.0064

please revert for any clarification required

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