Question

Six-week-old babies in the US consume a mean µ of 15.6 ounces of milk per day, with a standard deviation ? of 2 ounces Complete the following.

a) Describe the shape and horizontal scaling on the graph of the distribution for amount of milk consumed by the population of all US six-week-old babies.

b) Find the probability that the amount of milk consumed by of a randomly selected baby from this group will be less than 14 ounces---that is, find P(x < 14). Based upon your result, state whether or not it is unusual to randomly select such a baby from this group, and explain why you chose "unusual" or "not unusual" as your answer.

c) Suppose all possible samples of size 40, taken from the population of all six-week-old babies in the US, are drawn and the mean amount of milk consumed per day is found for each resulting sample. Describe the shape and scaling on the graph of the resulting sampling distribution for the sample mean values. Hint: Apply the Central Limit Theorem!

d) Find the probability that the mean milk consumed by a randomly selected sample of 40 six-week-old babies in the US will be less than 14 ounces---that is, find P(x-bar < 14). Based upon your result, state whether or not it is unusual to randomly select a sample of 40 such babies whose mean milk consumption per week is less than 14 ounces, and explain why you chose "unusual" or "not unusual" as your answer.

e) Find the probability that the mean milk consumption per week from a sample of 40 such babies will be between 15 and 16 ounces.

Answer 1

(a)

Assuming milk consumption is normally distributed.

Following is the curve:

b)

The z-score for X = 14 is

$z=\frac{x-\mu}{\sigma}=\frac{14-15.6}{2}=-0.8$

The required probability is:

P(X< 14) = P(z < -0.8) = 0.2119

Since this probability is greater than 0.05 so thsi is not unusual.

(C)

The sampling distribution of sample mean will be approximately normal distribution is:

$\mu_{\bar{x}}=\mu=15.6$

and standard deviation

$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{40}}=0.3162$

(d)

The z-score for $\bar{x}=14$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{14-15.6}{0.3162}=-5.06$

The required probability is:

$P(\bar{x}< 14) = P(z < -5.06) = 0.0000$

Since this probability is greater than 0.05 so thsi is not unusual.

e)

The z-score for $\bar{x}=15$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{15-15.6}{0.3162}=-1.90$

The z-score for $\bar{x}=16$ is

$z=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\frac{16-15.6}{0.3162}=1.27$

The probability that the mean milk consumption per week from a sample of 40 such babies will be between 15 and 16 ounces is

$P(15<\bar{x}<16)=P(-1.90<z<1.27)=0.8692$