Sampling Distributions

Consider exercise 6.38 on page 399 of your textbook.

Plastic bags used for packaging produce are manufactured so that the breaking strength of the bag is normally distributed with a mean of 5 pounds per square inch and a standard deviation of 1.5 pounds per square inch. A sample of 25 bags is selected.

So we have that the breaking strength of the bags is normal with $\mu = 5$ lbs/in² and $\sigma = 1.5$ lbs/in². Also, n = 25.

(a)(1) What is the probability that the average breaking strength is between 5 and 5.5 pounds per square inch?

$P(5 < \overline{X} < 5.5) = P(\frac{5 - 5}{1.5/\sqrt{25}} < Z < \frac{5.5 - 5}{1.5/\sqrt{25}}) =$
P(0 < Z < 1.6667) = .45220967

Note that if we want to specify $\mu$ and $\sigma$ in the normalcdf command, we actually have to plug in $\mu = 5$ and $\sigma / \sqrt{n} = 1.5 / \sqrt{25} = 0.3$. You should always draw a curve, and give it some thought. Use your common sense to decide if the answer the calculator comes up with is reasonable.

(a)(2) What is the probability that the average breaking strength is between 4.2 and 4.5 pounds per square inch?

$P(4.2 < \overline{X} < 4.5) = P(\frac{4.2 - 5}{1.5/\sqrt{25}} < Z < \frac{4.5 - 5}{1.5/\sqrt{25}}) =$
P(-2.6667 < Z < -1.6667) = .04395991

(a)(3) What is the probability that the average breaking strength is less than 4.6 pounds per square inch?

$P(\overline{X} < 4.6) = P(Z < \frac{4.6 - 5}{1.5/\sqrt{25}}) =$
P(Z < -1.3333) = .09121128

(b) Between what two values symmetrically distributed around the mean will 95% of the average breaking strengths be?

The values asked for here will be the 2.5^th percentile, and the 97.5^th percentile, since 95% of the average breaking strengths will be between that.
We need the Z quantile corresponding to the .025 tail area.

So the $\overline{X}$ value, once we "un-standardize", is:

$-1.95996 = \frac{\overline{X} - 5}{1.5/\sqrt{25}}$
$-0.587989 = \overline{X} - 5$
X = 4.412

And from the symmetry of the normal curve, the upper value will be

$1.95996 = \frac{\overline{X} - 5}{1.5/\sqrt{25}}$
$0.587989 = \overline{X} - 5$
X = 5.588

So 95% of the average breaking strengths will fall between 4.412 and 5.588.

(c) What will your answers be to (a) and (b) if the standard deviation is 1.0 pound per square inch?

Do this part on your own!
Here are the answers:

(a)(1) .49379 (a)(2) .00617799 (a)(3) .002275 (b) (4.608, 5.392)

Consider exercise 6.46 on page 402-403 of your textbook.

Historically, 93% of the deliveries of an overnight mail service arrive before 10:30 the following morning. Random samples of 500 deliveries are selected.

So we have a "historic" (read: population) proportion of p = .93. Also, n = 500.

(a) What proportion of the samples will have between 93% and 95% of the deliveries arriving before 10:30 the following morning?

$P(.93 < p_{s} < .95) = P(\frac{.93 - .93}{\sqrt{\frac{.93(.07)}{500}}} < Z < \frac{.95 - .93}{\sqrt{\frac{.93(.07)}{500}}}) =$
P(0 < Z < 1.7528) = .46017915

Note that if we want to specify "$\mu$" and "$\sigma$" in the normalcdf command, we actually have to plug in p =.93 and $\sqrt{\frac{.93(.07)}{500}}$.

(b) What proportion of the samples will have more than 95% of the deliveries arriving before 10:30 the following morning?

$P(p_{s} > .95) = P(Z > \frac{.95 - .93}{\sqrt{\frac{.93(.07)}{500}}}) =$
P(Z > 1.7528) = .03982085

(c) If samples of size 1000 are selected, what will your answers be in (a) and (b)?

Do this part on your own!
Here are the answers:

(a) .49340852 (b) .00659148

(d) Which is more likely to occur - more than 95% of the deliveries in a sample of 500 or less than 90 before 10:30 the following morning?

We will answer the question of which is more "likely" by simply calculating the probability of each event:

P(p_s > .95) = .03982085 (from part b)
$P(p_{s} < .90) = P(Z < \frac{.90 - .93}{\sqrt{\frac{.93(.07)}{1000}}}) =$
P(Z < -3.7182) = .00010036

Therefore, more than 95% of the deliveries in a sample of 500 arriving before 10:30 the following morning is more likely to occur, since it has a higher probability.