The weights of bowling balls are normally distributed with mean 11.5 pounds and standard deviation 2.7 pounds. A sample of 36 bowling balls is selected. What is the probability that the average weight of the sample is less than 11.03 pounds?
Solution :
Given that,
mean =
= 11.5
standard deviation =
= 2.7
n = 36
= 11.5
=
/
n = 2.7 /
36=0.45
P(
< 11.03) = P[(
-
) /
< (11.03-11.5) /0.45 ]
= P(z < -1.04)
Using z table
= 0.1492
probability= 0.1492
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