his Question: 3 pts 3 0f 25 (1 complete) This Quiz: 60 pts possible Suppose the
ID: 3053239 • Letter: H
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his Question: 3 pts 3 0f 25 (1 complete) This Quiz: 60 pts possible Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u 257 days and standard deviaton o 22 days Complete parts (a) through (1) below. (a) What is the probability that a randomly selected pregnancy lasts less than 250 days? The probabilit that a randomly selected pregnancy lasts less than 250 days is approximately(Round to four decimal places as needed) Interpret this probability. Select the correct choice below and fill in the answer box within your choice Round to the nearest integer as needed.) O A. I100 pregnant indrviduals were selected independently from this population, we would expect pregnancies to last exactly 250 days. OB. If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last more than 250 days. ? ? 100 regnant i duals were selected independently from this population, we would expect pregnancies to last less than 250 days (b) Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. The sampling distribution of x is Round to four decimal places as needed ) withK"I and%"DExplanation / Answer
a)
mean = 257 , std.dev = 22
p(x <250)
z = (x -mean) /s
= ( 250 -257)/22
= -0.3182
P(x <250) = P(z <-0.3182 ) = 0.3752 by using standard normal table
expected number = 0.3752*100 = 37.32 i.e. 37
Option C : 37 pregnancies to last less than 250 days.
b)
The sampling distribution of xbar is = 257 , std.deviation = s/sqrt(n) = 22/sqrt(20) = 4.9193
c)
n = 20
p(x <250)
z = (x -mean) /s
= ( 250 -257)/(22/sqrt(20))
= -1.4230
P(x <250) = P(z < -1.4230 ) = 0.0774 by using standard normal table
expected number = 100*0.0774 = 7.74 i.e. 8
Interpretation Correct answer is Option A) : 8 samples to have sample mean of 250 days or less
d)
n = 50
p(x <250)
z = (x -mean) /s
= ( 250 -257)/(22/sqrt(50))
= -2.2499
P(x <250) = P(z < -2.2499) = 0.0122 by using standard normal table
expected number = 100*0.0122 = 1.22 i.e. 1
Interpretation Correct answer is OPtion B) : 1 sample to have mean of 250 days or less.
f)
p((257 -12) < x < (257+12))
P(245 < x < 269)
= P((245 - 257) / (22/sqrt(20) < z < (269 - 257) /(22/sqrt(20))
= P(-2.4393 < z < 2.4393)
P(245 < x < 269)
= P(-2.4393 < z < 2.4393)
= 0.9853 ..by using standard normal table
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