Answer:
Given:
X = 149
Mean, u = 146
Standard deviation [tex] \sigma [/tex] = 2.3
a) Let's find the zscore at x = 149
[tex] z = \frac{x - u}{\sigma} [/tex]
[tex] z = \frac{149 - 146}{2.3} = 1.3043 [/tex]
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
[tex] \sigma [/tex] = [tex] \sqrt{npq} [/tex]
[tex] = \sqrt{16*0.9039*0.0961} = 1.1789 [/tex]
By continuity,
P(x≤15) = P(x<15.5)
At x = 15.5
[tex] z = \frac{x - u}{\sigma} [/tex]
[tex] z = \frac{15.5 - 14.4624}{1.1789} = 0.8801[/tex]
Therefore
P(x≤15) = P(x<15.5) =
P(z<0.8801) =
NORMSDIST(0.8801) = 0.8106
c) Given:
X = 147
P(X>147) =
[tex] z = \frac{x - u}{\sigma/ \sqrt{n}} [/tex]
[tex] z = \frac{147 - 146}{2.3/ \sqrt{16}} = 1.7391[/tex]
P(X>147) = P(z > 1.7391)
NORMDIST(1.7391) = 0.9590
P(X>147) = 1 - 0.9590
= 0.041
