Question

Columbia manufactures bowling balls with a mean weight of 14.5 pounds and a standard deviation of 2.7 pounds. A bowling ball is too heavy to use and is discarded if it weighs over 16 pounds. Assume that the weights of bowling balls manufactured by Columbia are normally distributed.

(Round probabilities to four decimals)

a) What is the probability that a randomly selected bowling ball is discarded due to being too heavy to use? _________

b) The lightest 6% of the bowling balls made are discarded due to the possibility of defects. A bowling ball is discarded for being too light if it weighs under what specific weight? (Round weight to two decimals)

_________ pounds

c) What is the probability that a randomly selected bowling ball will be discarded for being either too heavy or too light? ____________

Answer 1

Solution:

We are given that weights are normally distributed.

Mean = 14.5

SD = 2.7

a) What is the probability that a randomly selected bowling ball is discarded due to being too heavy to use?

Here, we have to find P(X>16)

P(X>16) = 1 – P(X<16)

Z = (X – mean) / SD

Z = (16 - 14.5)/2.7

Z = 0.555556

P(Z<0.555556) = P(X<16) = 0.710743

(By using z-table)

P(X>16) = 1 – P(X<16)

P(X>16) = 1 – 0.710743

P(X>16) = 0.289257

Required probability = 0.2893

b) The lightest 6% of the bowling balls made are discarded due to the possibility of defects. A bowling ball is discarded for being too light if it weighs under what specific weight? (Round weight to two decimals)

Z score for lowest 6% is given as below:

P(X<x) = 0.06

Z = -1.55477

(by using z-table)

X = Mean + Z*SD

X = 14.5 + (-1.55477)*2.7

X = 10.30212

Required weight = 10.30 Pounds

c) What is the probability that a randomly selected bowling ball will be discarded for being either too heavy or too light?

Here, we have to find P(X<10.30) + P(X>16)

P(X<10.30) + P(X>16) = 0.06 + 0.2893

P(X<10.30) + P(X>16) = 0.3493

Required probability = 0.3493