Question

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

a.What is the distribution for the weights of one 25-pound lifting weight? What is the mean and standard deviation?

b.What is the distribution for the mean weight of 100 25-pound lifting weights?

c.Find the probability that the mean actual weight for the 100 weights is less than 24.9.

d.Find the 90th percentile for the mean weight for the 100 weights.

e.Find the probability that the mean actual weight for the 100 weights is greater than 25.2.

Answer 1

a)

distribution for the weights of one 25-pound lifting weight is uniform~ U(24,26)

mean μ=(a+b)/2 =

25

standard deviation σ=(b-a)/√12=

0.5774

b)

distribution for the mean weight of 100 25-pound lifting weights will be approximately normal

~ N(25,0.5774/√100) =~ N(25,0.0577)

c)

probability =P(X<24.9)=(Z<24.9-25)/0.058)=P(Z<(-1.7321)=0.0416

d)

for 90th percentile critical value of z=		1.282
therefore corresponding value=mean+z*std deviation=			25.0740

e)

probability =P(X>25.2)=P(Z>(25.2-25)/0.058)=P(Z>3.46)=1-P(Z<3.46)=1-0.9997=0.0003