Question

8. The lengths of pregnancies are normally distributed with a mean of 267 days and a standard deviation of 15 days. (a) Find the probability that an individual woman has a pregnancy shorter than 259 days. (b) If 36 women are randomly selected, find the probability that they have a mean preg- nancy shorter than 259 days. (c) There should be a difference in your method for the previous two questions. Explain what you did differently for each problem and explain WHY your answers are different. (d) Find the probability that an individual woman has a pregnancy longer than 295 days. (e) Find the probability that an individual woman has a pregnancy between 259 and 295 days. (f) What is the cutoff number of days for a pregnancy for the top 15% of women? (g) What is the cutoff number of days for a pregnancy for the bottom 25% of women?

Answer 1

Solution :

Let X be a random variable which represents the length of pregnancies.

Given that, X ~ N(267, 15²)

μ = 267 days and  σ = 15 days

a) We have to find P(X < 259).

We know that if X ~ N(μ, σ²) then, $\large Z = \frac{X-\mu}{\sigma}\sim N(0,1)$

$\large \therefore P(X < 259) = P\left(\frac{X-\mu}{\sigma} < \frac{259-\mu}{\sigma} \right)$

::P(X < 259) = P(Z < 259 – 267 15

.: P(X < 259) = P(Z < -0.5333

Using "pnorm" function of R we get, P(Z < -0.5333) = 0.2969

.: P(X < 259) = 0.2969

Hence, the required probability is 0.2969.

b) We have to find P(x̄ < 259).

We know that if X ~ N(μ, σ²) then, x̄ ~ N(μ, σ²/n).

And if x̄ ~ N(μ, σ²/n) then,

Z T- o/vn N(0,1)

Sample size (n) = 36

$\large \therefore P(\bar{x} < 259) = P\left ( \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} <\frac{259-\mu}{\sigma/\sqrt{n}} \right )$

$\large \therefore P(\bar{x} < 259) = P\left (Z <\frac{259-267}{15/\sqrt{36}} \right )$

$\large \therefore P(\bar{x} < 259) = P\left (Z < -3.2 \right )$

Using "pnorm" function of R we get, P(Z < -3.2) = 0.0007

$\large \therefore P(\bar{x} < 259) = 0.0007$

Hence, the required probability is 0.0007.

c) In part (a) we have used the distribution of population of pregnancies, whereas in the part (b) we have used the sampling distribution of sample means to obtain the probability.

The answers are different because population distribution has different standard deviation than the standard deviation of the sampling distribution of sample means.

d) We have to find P(X > 295).

We know that if X ~ N(μ, σ²) then, $\large Z = \frac{X-\mu}{\sigma}\sim N(0,1)$

$\large \therefore P(X > 295) = P\left(\frac{X-\mu}{\sigma} > \frac{295-\mu}{\sigma} \right)$

$\large \therefore P(X > 295) = P\left(Z > \frac{295-267}{15} \right)$

$\large \therefore P(X > 295) = P\left(Z > 1.8667 \right)$

Using "pnorm" function of R we get, P(Z < 1.8667) = 0.0310

$\large \therefore P(X > 295) = 0.0310$

Hence, the required probability is 0.0310.