question archive Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 131 days μ=131 days and standard deviation sigma equals 8 days σ=8 days
Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 131 days μ=131 days and standard deviation sigma equals 8 days σ=8 days
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Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 131 days
μ=131 days and standard deviation sigma equals 8 days
σ=8 days. Complete parts? (a) through? (f) below.
?(a) What is the probability that a randomly selected pregnancy lasts less than 128
128 ?days?
The probability that a randomly selected pregnancy lasts less than 128
128 days is approximately
0.352
0.352. ?(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.)
A.
If 100 pregnant individuals were selected independently from this? population, we would expect
35
35 pregnancies to last less than 128
128 days.
Your answer is correct.
B.
If 100 pregnant individuals were selected independently from this? population, we would expect
nothing
pregnancies to last more than 128
128 days.
C.
If 100 pregnant individuals were selected independently from this? population, we would expect
nothing
pregnancies to last exactly 128
128 days.
?(b) Suppose a random sample of 19
19 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies.
The sampling distribution of x overbar
x is normal
with mu Subscript x overbar
μxequals
=
131
131 and sigma Subscript x overbar
σxequals
=
1.8353
1.8353.
?(Round to four decimal places as? needed.)
?(c) What is the probability that a random sample of 19
19 pregnancies has a mean gestation period of 128
128 days or? less?
The probability that the mean of a random sample of 19
19 pregnancies is less than 128
128 days is approximately
nothing
?(Round to four decimal places as? needed.)
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(a) P(X < 128) = 0.3538
A.
If 100 pregnant individuals were selected independently from this? population, we would expect
35 pregnancies to last less than 128 days.
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(b) for sample size (n) = 19
mean of the sampling distribution ( x ) = 131
standard deviation ( sd )= 8/ sqrt ( 19 ) =1.8353
p(x <= 128) = 0.0511
if assumed the continuity correction,
p(x <= 128) = 0.0866
Step-by-step explanation
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 131
standard deviation ( sd )= 8
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(a)
P(X < 128) = (128-131)/8
= -3/8= -0.375
= P ( Z <-0.375) from standard normal table
= 0.3538
A.
If 100 pregnant individuals were selected independently from this? population, we would expect
35 pregnancies to last less than 128 days.
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(b) for sample size (n) = 19
mean of the sampling distribution ( x ) = 131
standard deviation ( sd )= 8/ sqrt ( 19 ) =1.8353
p(x <= 128) = (128-131)/8/ sqrt ( 19 )
= -3/1.8353= -1.6346
= p ( z <-1.6346) from standard normal table
= 0.0511
if assumed the continuity correction,
p(x <= 128) = p(x < 128.5) = (128.5-131)/8/ sqrt ( 19 )
= -2.5/1.8353= -1.3622
= p ( z <-1.3622) from standard normal table
= 0.0866
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