Columbia manufactures bowling balls with a mean weight of 14.6 pounds and a stan
ID: 3336699 • Letter: C
Question
Columbia manufactures bowling balls with a mean weight of 14.6 pounds and a standard deviation of 1.6 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?
Explanation / Answer
Solution:
We are given that weights are normally distributed.
Mean = 14.6
SD = 1.6
a) What is the probability that a randomly selected bowling ball is discarded due to being too heavy to use?
Solution:
We have to find P(X>16)
P(X>16) = 1 – P(X<16)
Z = (X – Mean) / SD
Z = (16 – 14.6) / 1.6
Z = 0.875
P(Z< 0.875) = P(X<16) = 0.809213047
P(X>16) = 1 – P(X<16)
P(X>16) = 1 – 0.809213047
P(X>16) = 0.190786953
Required probability = 0.1908
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
Solution:
We have formula
X = Mean + Z*SD
Z value for lightest 6% is given as -1.554773595 (by using z-table or excel)
X = 14.6 + (-1.554773595)*1.6
X = 14.6 – 1.554773595*1.6
X = 12.11236225
Specific weight = 12.11 pounds
c) What is the probability that a randomly selected bowling ball will be discarded for being either too heavy or too light?
Solution:
We have to find P(X<12.11) + P(X>16)
From part a, we have
P(X>16) = 0.190786953
Now, we have to find P(X<12.11)
Z = (12.11 – 14.6) / 1.6
Z = -1.55625
P(Z< -1.55625) = P(X<12.11) = 0.059824328
P(X<12.11) + P(X>16) = 0.190786953 + 0.059824328
P(X<12.11) + P(X>16) = 0.250611281
Required probability = 0.2506