Question

A regulation baseball can weigh no more than 149 grams. A factory produces baseballs with weights that are normally distributed with a mean of 146 grams and a standard deviation of 2.3 grams.

(a) If a baseball produced by the factory is randomly selected, what is the probability that it is within regulation weight?
(b) The baseballs are shipped in boxes of 16. What is the probability that at least 15 of the 16 baseballs in a pack are within regulation weight?
(c) The factory will not ship a box of 16 if the average weight of the baseballs in the box exceeds 147 grams. What is the probability that a pack of 16 baseballs would have an average weight of more than 147 grams?

Answer 1

Given:

X = 149

Mean, u = 146

Standard deviation
`\sigma` = 2.3

a) Let's find the zscore at x = 149

`z = (x - u)/(\sigma)`

`z = (149 - 146)/(2.3) = 1.3043`

From the standard normal table

NORMSDIST(1.3043) = 0.9039

Therefore

P(x>149) = P(z> 1.3043)

= 1 - 0.90393

= 0.0961

The probability that the baseball is within regulation weight is

1 - 0.0961 = 0.9039

b) Let's use binomial distribution,

Given :

n = 16

p = 0.9039

q = 1-0.9039 = 0.0961

Mean = np

np = 16 * 0.9039 = 14.4624

`\sigma` =
`√(npq)`

`= √(16*0.9039*0.0961) = 1.1789`

By continuity,

P(x≤15) = P(x<15.5)

At x = 15.5

`z = (x - u)/(\sigma)`

`z = (15.5 - 14.4624)/(1.1789) = 0.8801`

Therefore

P(x≤15) = P(x<15.5) =

P(z<0.8801) =

NORMSDIST(0.8801) = 0.8106

c) Given:

X = 147

P(X>147) =

`z = (x - u)/(\sigma/ √(n))`

`z = (147 - 146)/(2.3/ √(16)) = 1.7391`

P(X>147) = P(z > 1.7391)

NORMDIST(1.7391) = 0.9590

P(X>147) = 1 - 0.9590

= 0.041