Mathematics, 18.07.2020 22:01 AgentPangolin

Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 247 days and standard deviation sigma equals 16 days. Complete parts (a) through (f) below.

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Answer from: safiyyahrahman6907

The answer is given below

Step-by-step explanation:

a) What is the probability that a randomly selected pregnancy lasts less than 242 days

First we have to calculate the z score. The z score is used to determine the measure of standard deviation by which the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

Given that Mean (μ) = 247 and standard deviation (σ) = 16 days. For x < 242 days,

$z=\frac{x-\mu}{\sigma}=\frac{242-247}{16}=-0.31$

From the normal distribution table, P(x < 242) = P(z < -0.3125) = 0.3783

(b) Suppose a random sample of 17 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies.

If a sample of 17 pregnancies is obtained, the new mean $\mu_x=\mu=247,$ the new standard deviation: $\sigma_x=\sigma/\sqrt{n} =16/\sqrt{17} =3.88$

c) What is the probability that a random sample of 17 pregnancies has a mean gestation period of 242 days or less

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{242-247}{16/\sqrt{17} }=-1.29$

From the normal distribution table, P(x < 242) = P(z < -1.29) = 0.0985

d) What is the probability that a random sample of 49 pregnancies has a mean gestation period of 242 days or less?

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{242-247}{16/\sqrt{49} }=-2.19$

From the normal distribution table, P(x < 242) = P(z < -2.19) = 0.0143

(e) What might you conclude if a random sample of 49 pregnancies resulted in a mean gestation period of 242 days or less?

It would be unusual if it came from mean of 247 days

f) What is the probability a random sample of size 2020 will have a mean gestation period within 11 days of the mean

For x = 236 days

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{236-247}{16/\sqrt{20} }=-3.07$

For x = 258 days

$z=\frac{x-\mu}{\sigma/\sqrt{n} }=\frac{258-247}{16/\sqrt{20} }=3.07$

From the normal distribution table, P(236 < x < 258) = P(-3.07 < z < 3.07) = P(z < 3.07) - P(z < -3.07) =0.9985 - 0.0011 = 0.9939

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so 1/4 f(x)

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