Assume that women’s heights are normally distributed with a mean given by µ = 63.3 in, and a standard devia?
Assume that women’s heights are normally distributed with a mean given by µ = 63.3 in, and a standard deviation given by ơ = 2.8 in.
a. If one woman is randomly selected, find the probability that her height is less than 65in.
b. If 46 women are randomly selected, find the probability that they have a mean height less than 65in.
a.The probability is approximately _________?
(Round to four decimal places as needed)
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65-63.3/2.8=.607
Pr[Z<.607]=.7281
65-63.3/2.8/sqrt(46)=4.12
Pr[Z<4.12]=.999981
(The standard deviation of the sample mean
is ơ/sqrt(n))
a) z₆₅ = (65-63.3)/2.8 ≈ .61
When you look up .61 in a z table, you see .7257 in the column. Therefore, the probability that the woman's height is less than 65 inches is 72.57%.