Question

Columbia manufactures bowling balls with a mean weight of 14.5 pounds and a standard deviation of 2.2 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 decimal places. a) What is the probability that a randomly selected bowling ball is discarded due to being too heavy to use? 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? (Round weight to two decimal places.) 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

Given: = 14.5, = 2.2

To find the probability, we need to find the Z scores first.

Z = (X - )/

We can find p values given the z scores by either using normal distribution tables, online calculators or by using the excel formula NORMSDIST(z score).

Similarly we can find the z scores from the given p values, by using the normal distribution tables, online calculators or by using the excel formula NORMSINV(p value). In the calculation below, I have used the excel formulas for more precision of values.

Sometimes a normal distribution table is provided and you may be required to use it instead of calculators. After part (c) I have re-evaluated the z scores and probabilities for (a) (b) and (c), with values from z tables.

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(a) P(The ball is discarded since it is too heavy to use) = P(X > 16)

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

For P( X < 16), Z = (16 – 14.5)/2.2 = 0.6818

The probability for P(X < 16) using NORMSDIST(0.6818) = 0.7523

Therefore the required probability P(X > 16) = 1 – 0.7523 = 0.2477

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(b) The lightest 5% are discarded. Therefore this value is below the mean, and we need to find the value of the weight of the ball. Therefore P(X < x) = 0.05.

The Z value at p = 0.05 (using NORMSINV(0.05)) = -1.645

Therefore -1.645 = (X - 14.5)/2.2. Solving for X, we get X = (-1.645 * 2.2) + 14.5 = 10.881

Rounding to 2 decimal places, the weight under which a bowling ball is discarded = 10.88 pounds.

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(c) P(A ball is discarded for being too heavy or too light) = P(X > 16 pounds) + P(X < 10.88 pounds)

= 0.2477 + 0.0500 = 0.2977

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In Case you are using z tables. You may see that the difference in values is very small.

(a) Z = 0.6818 0.68

The p value for X < 16 from the normal tables is 0.7518

Therefore P(X > 16) = 1 - 0.7518 = 0.2482

(b) This value is the same as calculated. = 10.88 pounds, as z tables and NORMINV(0.05) is the same for both = -1.645.

(c) This value = P(X > 16) + P(X < 10.88) = 0.2482 + 0.0500 = 0.2982

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