sampling distribution

Question

A quality control company was hired to study the length of meter sticks produced by a certain company. The team carefully measured the length of many many meter sticks, and the distribution seems to be slightly skewed to the right with a mean of 99.9 cm and a standard deviation of 0.19 cm.

a) What is the probability of finding a meter stick with a length of more than 100.11 cm?

b) What is the probability of finding a group of 32 meter sticks with a mean length of less than 99.88 cm?

c) What is the probability of finding a group of 26 meter sticks with a mean length of more than 99.93 cm?

d) What is the probability of finding a group of 42 meter sticks with a mean length of between 99.89 and 99.92 cm?

e) For a random sample of 38 meter sticks, what mean length would be at the 92nd percentile?

Answer 1

x = stick length N(99.9,0.19)

z = (x - mean)/std dev to transform x to standard normal distribution N(0,1)

a) P(x > 100.11) = P(z > (100.11 - 99.9)/0.19) = P(z > 1.11) = 1 - P(z < 1.11) = 1 - 0.8665 = 0.1335

xbar = mean stick length N(99.9,0.19)

z = (xbar - mean)/(std dev/sqrt(sample size) to transform xbar to standard normal distribution N(0,1)

b) P(xbar < 99.88) = P(z < (99.88 - 99.9)/(0.19/sqrt(32)) = P(z < -0.60) = 0.2743

c) P(xbar > 99.93) = P(z > (99.93 - 99.9)/(0.19/sqrt(26)) = P(z > 0.81) = 1 - P(z < 0.81) = 1 - 0.7910 = 0.209

d)_ P(99.89 < xbar < 99.92) = P((99.89-99.9)/(0.19/sqrt(42)) < z < P((99.92-99.9)/(0.19/sqrt(42)) = P(-0.34< z < 0.68) = P(z < 0.68) - P(z < -0.34) = 0.7517 - 0.3669 = 0.3848

e) 92nd percentile corresponds to a z-score of 1.45

using formula z = (xbar - mean)/(std dev/sqrt(sample size):

1.45 = (xbar - 99.9)/(0.19/sqrt(38))

1.45 * (0.19/sqrt(38)) + 99.9 = xbar

99.94 = XBAR