Question

NO HANDWRITTEN ANSWERS PLEASE

For the standard normal curve, find the z-score that corresponds to the 30th percentile.

Find the z-score for which 99% of the distribution’s area lies between –z and z.

Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height between 68 and 70 inches.

An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows.

Answer 1

For the standard normal curve, find the z-score that corresponds to the 30th percentile.

Answer:

Z-score = -0.5244

(by using z-table or excel)

Find the z-score for which 99% of the distribution’s area lies between –z and z.

Answer:

We are given middle area = 99%

So, area at both sides = 1% = 0.01

So, area at left side = 0.01/2 = 0.005

Z score corresponding the probability 0.005 is

Z = -2.57583

Z score for area at right side = 2.57583

Z = 2.57583

Assume that the heights of men are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. If 64 men are randomly selected, find the probability that they have a mean height between 68 and 70 inches.

µ = 69, σ = 2.8, n=64

We have to find P(68<Xbar<70) = P(Xbar<70) – P(Xbar<68)

z = (70 – 69)/[2.8/sqrt(64)] = 2.857143

P(Z< 2.857143) = P(Xbar<70) = 0.997863

Z = (68 - 69)/[2.8/sqrt(64)] = -2.85714

P(Z<-2.85714) = P(Xbar<68) = 0.002137

P(68<Xbar<70) = P(Xbar<70) – P(Xbar<68)

P(68<Xbar<70) = 0.997863 - 0.002137

P(68<Xbar<70) = 0.99573

Required probability = 0.99573

An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows.

We are given n=150, p=0.15, q=1- p = 0.85

We have to use normal approximation to binomial distribution.

We have to find P(X<20) = P(X<19.5) (by using continuity correction)

Mean = np = 150*0.15 = 22.5

SD = sqrt(npq) = sqrt(150*0.15*0.85) = 4.373214

Z = (X – mean) / SD

Z = (19.5 - 22.5) / 4.373214

Z = -0.68599

P(Z<-0.68599) = 0.246358

Required probability = 0.246358

[All probabilities are obtained from Z-table or excel]