11. Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean μ=286 days and standard deviation sigma equals σ=29 days.
Complete parts (a) through (f) below.
I cannot figure out part (f), please help.
(f) What is the probability a random sample of size 18 will have a mean gestation period within 11 days of the mean?
The probability that a random sample of size 18 will have a mean gestation period within 11 days of the mean is ______.
(Round to four decimal places as needed.)
Solution :
Given that ,
mean = = 286 days
standard deviation = = 29 days
n = 18
_{} = = 286
_{} = / n = 29 / 18 = 6.84
286 ± 11 = 275, 297
P(275 < < 297)
= P[(275 - 286) / 6.84 < ( - _{}) / _{} < (297 - 286) / 6.84 )]
= P(-1.61 < Z < 1.61)
= P(Z < 1.61) - P(Z < -1.61)
Using z table,
= 0.9463 - 0.0537
= 0.8926
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