The height of women ages 20-29 is normally dstnbuted, with a mean of 63 8 inches
ID: 3326634 • Letter: T
Question
The height of women ages 20-29 is normally dstnbuted, with a mean of 63 8 inches height less than 64.9 inches or are you more likely to select a sample of 27 women with a mean height less than 64.9 inches? Explain Assume = 2.7 inches Are you more likely to randomly select 1 woman with a Click the icon to view page 1 of the standard normal table Click the icon to view page 2 of the standard normal table What is the probability of randomly selecting 1 woman with a height less than 64.9 inches? (Round to four decimal places as needed ) What is the probability of selecting a sample of 27 women with a mean height less than 64 9 inc hes? (Round to four decimal places as needed ) fme Are than 64 9 inches? Choose the correct ans wer below C andtom celet nthon 6i4 nchee er seuee to selet a sngle af 27 somen A t is more likely to select a sample of 27 women with a mean height less than 649 inches because the sample of 27 has a lower probability O B. It is more likely to select 1 woman with a height less than 64 9 inc hes bec.ause the probability is lower O C. It is more likely to select 1 woman with a height less than 64 9 inches because the probability is higher O D. It is more likely to select a sample of 27 women with a mean height less than 64 9 inches because the sample of 27 has a higher probability Click to select your answeris)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/sqrt(n) ~ N(0,1)
mean of the sampling distribution ( x ) = 63.8
standard Deviation ( sd )= 2.7/ Sqrt ( 1 ) =2.7
sample size (n) = 1
P(X < 64.9) = (64.9-63.8)/2.7/ Sqrt ( 1 )
= 1.1/2.7= 0.4074
= P ( Z <0.4074) From Standard NOrmal Table
= 0.65815
b.
mean of the sampling distribution ( x ) = 63.8
standard Deviation ( sd )= 2.7/ Sqrt ( 27 ) =0.5196
sample size (n) = 27
P(X < 64.9) = (64.9-63.8)/2.7/ Sqrt ( 27 )
= 1.1/0.5196= 2.117
= P ( Z <2.117) From Standard NOrmal Table
= 0.98287
c.
option A
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