core: 0.67 of 1 pt 17 of 26 (15 complete) HYV Score: 54.49%, 14.17 of 26 6.5.7-T
ID: 2930113 • Letter: C
Question
core: 0.67 of 1 pt 17 of 26 (15 complete) HYV Score: 54.49%, 14.17 of 26 6.5.7-T Question Help The overhead reach distances of adult females are normally distributed with a mean of 202.5 cm and a standard deviation of 8 cm. a. Find the probability that an individual distance is greater than 215.90 cm b. Find the probability that the mean for 25 randomly selected distances is greater than 200.70 cm. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?Explanation / Answer
a)
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 202.5
standard Deviation ( sd )= 8
P(X > 215.9) = (215.9-202.5)/8
= 13.4/8 = 1.675
= P ( Z >1.675) From Standard Normal Table
= 0.047
b)
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 202.5
standard Deviation ( sd )= 8
sample size (n) = 25
P(X > 200.7) = (200.7-202.5)/8/ Sqrt ( 25 )
= -1.8/1.6= -1.125
= P ( Z >-1.125) From Standard Normal Table
= 0.8697
c)
According to the Central Limit Theorem (CLT) we will first illustrate the result. In order for the result of the CLT to hold, the sample must be sufficiently large (n > 30). Again, there are two exceptions to this. If the population is normal, then the result holds for samples of any size (i..e, the sampling distribution of the sample means will be approximately normal even for samples of size less than 30).
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