Question 1064321: Assume that women's heights are normally distributed with a mean given by mu equals 62.4 in, and a standard deviation given by sigma equals 1.8 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 30 women are randomly selected, find the probability that they have a mean height less than 63 in.
(a)The probability is approximately . 6293
.
(Round to four decimal places as needed.)
(b) The probability is approximately
.
I have (A) I need help with Part B.
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Assume that women's heights are normally distributed with a mean given by mu equals 62.4 in, and a standard deviation given by sigma equals 1.8 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 30 women are randomly selected, find the probability that they have a mean height less than 63 in.
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For samples of size 30 the standard deviation becomes 1.8/sqrt(30).
z(63) = (63-62.4)/(1.8/sqrt(30)) = 1.8257
P(x-bar< 63) = P(z < 1.8257) = normalcdf(-100,1.8257) = 0.9661
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a)The probability is approximately 0.6293
(b) The probability is approximately 0.9661
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Cheers,
Stan H.
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Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! (B) This portion of the problem deals with the sampling distribution of the mean
:
our sample size is 30, n = 30
:
we know the following information
:
the sample mean is equal to the population mean
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the sample standard deviation is call the standard error(SE) and
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SE = population standard deviation / square root(n)
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SE = 1.8 / square root(30) = 0.3286
:
because the sample size > or = 30, we can use the normal distribution for the sample
:
z-score (X < 63 ) = ( 63 - 62.4 ) / 0.3286 = 1.8259
:
Probability ( X < 63 ) = 0.9664
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