In: Statistics and Probability
Assume that? women's heights are normally distributed with a mean given by mu equals 64.6 in?, and a standard deviation given by sigma equals 1.9 in. ?(a) If 1 woman is randomly? selected, find the probability that her height is less than 65 in. ?(b) If 50 women are randomly? selected, find the probability that they have a mean height less than 65 in. ?(?a) The probability is approximately nothing. ?(Round to four decimal places as? needed.) ?(b) The probability is approximately nothing. ?(Round to four decimal places as? needed.)
Solution:
Given that,
mean = = 64.6
standard deviation = = 1.9
A ) p ( x < 65 )
= p ( x - / ) < ( 65 - 64.6 / 1.9 )
= p ( z < 0.4 / 1.9 )
= p ( z < 0.21)
Using z table
= 0.5832
Probability = 0.5832
B ) n = 50
So,
= 64.6
= ( /n) = (1.9 / 50 ) = 0.2687
p ( < 65 )
= p ( - / ) < ( 65 - 64.6 / 0.2687 )
= p ( z < 0.4 / 0.2687 )
= p ( z < 1.49)
Using z table
= 0.9319
Probability = 0.9319