Columbia manufactures bowling balls with a mean weight of 15.1 pounds and a stan
ID: 3335218 • Letter: C
Question
Columbia manufactures bowling balls with a mean weight of 15.1 pounds and a standard deviation of 2.1 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 7% 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
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 15.1
standard Deviation ( sd )= 2.1
a.
the probability that a randomly selected bowling ball is discarded due to being too heavy to use
P(X > 16) = (16-15.1)/2.1
= 0.9/2.1 = 0.4286
= P ( Z >0.4286) From Standard Normal Table
= 0.3341
b.
The lightest 7% 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
P ( Z < x ) = 0.07
Value of z to the cumulative probability of 0.07 from normal table is -1.48
P( x-u/s.d < x - 15.1/2.1 ) = 0.07
That is, ( x - 15.1/2.1 ) = -1.48
--> x = -1.48 * 2.1 + 15.1 = 12
c.
the probability that a randomly selected bowling ball will be discarded for being either too heavy or too lightTo find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 12) = (12-15.1)/2.1
= -3.1/2.1= -1.4762
= P ( Z <-1.4762) From Standard Normal Table
= 0.0699
P(X > 16) = (16-15.1)/2.1
= 0.9/2.1 = 0.4286
= P ( Z >0.4286) From Standard Normal Table
= 0.3341
P( X < 12 OR X > 16) = 0.0699+0.3341 = 0.4041