Question

Columbia manufactures bowling balls with a mean weight of 14.6 pounds and a standard deviation of 2.4 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.

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

The lightest 5% 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) pound

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: Columbia manufactures bowling balls with a mean weight of 14.6 pounds and a standard deviation of 2.4 pounds.

That is: and

the weights of bowling balls manufactured by Columbia are normally distributed.

A bowling ball is too heavy to use and is discarded if it weighs over 16 pounds.

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

That is we have to find:

P( X > 16 )= ..............?

Thus find z score for x = 16

Thus we get:

P( X > 16) = P( Z > 0.58)

P( X > 16) = 1 - P( Z < 0.58)

Look in z table for z = 0.5 and 0.08 and find corresponding area.

P( Z < 0.58) =0.7190

Thus

P( X > 16) = 1 - P( Z < 0.58)

P( X > 16) = 1 - 0.7190

P( X > 16) = 0.2810

Thus the probability that a randomly selected bowling ball is discarded due to being too heavy to use is 0.2810

**Part b) The lightest 5% 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?

That is we have to find x value such that:

P(X < x ) = 5%

P( X < x ) = 0.05

Look in z table for Area =0.0500 or its closest area and find z value.

Area 0.0500 is in between 0.0495 and 0.0505 and both the area are at same distance from 0.0500

Thus we look for both area and find both z values

Thus Area 0.0495 corresponds to -1.65 and 0.0505 corresponds to -1.64

Thus average of both z values is : ( -1.64+ - 1.65) / 2 = -1.645

Thus Z = -1.645

we use following formula:

pounds.

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

P( Ball is too heavy or too light ) =P( X > 16 ) + P( X < 10.65)

P( Ball is too heavy or too light ) = 0.2810 + 0.05

**P( Ball is too heavy or too light ) =
0.3310**

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