In: Statistics and Probability
The overhead reach distances of adult females are normally distributed with a mean of
197.5 cm
and a standard deviation of
8.9 cm
a. Find the probability that an individual distance is greater than
210.00
cm.
b. Find the probability that the mean for
20
randomly selected distances is greater than 195.70 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
SOLUTION:
From given data,
The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm and a standard deviation of 8.9 cm
a. Find the probability that an individual distance is greater than 210.00 cm.
b. Find the probability that the mean for 20 randomly selected distances is greater than 195.70 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
Where,
mean = = 197.5
standard deviation = = 8.9
Z = (X - ) / = (X -197.5 ) / 8.9
a. Find the probability that an individual distance is greater than 210.00 cm.
P(X > 210.00) = P( (X - ) / > (210.00 -197.5 ) / 8.9)
P(X > 210.00) = P( Z > (210.00 -197.5 ) / 8.9)
P(X > 210.00) = P( Z > 12.5 / 8.9)
P(X > 210.00) = P( Z > 1.40)
P(X > 210.00) = 1-P( Z < 1.40)
P(X > 210.00) = 1-0.91924 (from z table)
P(X > 210.00) = 0.08076
b. Find the probability that the mean for 20 randomly selected distances is greater than 195.70 cm.
Where ,
n = 20
= / sqrt(n) = 8.9 / sqrt(20) = 1.9901004
= = 197.5
Z = ( - ) / () = ( - 197.5) / (1.9901004)
P( > 195.70 ) = P(( - ) / () > (195.70 -197.5 ) / 1.9901004)
P( > 195.70 ) = P( Z > (195.70 -197.5 ) / 1.9901004)
P( > 195.70 ) = P( Z > -1.8 / 1.9901004)
P( > 195.70 ) = P( Z > -0.90)
P( > 195.70 ) = 1-P( Z < -0.90)
P(X > 195.70 ) = 1-0.18406 (from z table)
P(X > 195.70 ) = 0.81594
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
Because the original population has a normal distribution, the distribution of the sample mean is normal for any sample size.