Question

Question 1

The average math SAT score is 511with a standard deviation of 119. A particular high school claims that its students have unusually high math SAT scores. A random sample of 50 students from this school was​ selected, and the mean math SAT score was 555. Is the high school justified in its​ claim? Explain.

▼ Pick one

No

Yes

​because the​ z-score (what is the z score)

​(?​)

is

▼ pick one

not unusual

unusual

since it

▼ pick one

lies

does not lie

within the range of a usual​ event, namely within

▼ pick one

1 standard deviation

2 standard deviations

3 standard deviations

of the mean of the sample means

Question 2

The lengths of lumber a machine cuts are normally distributed with a mean of 102 inches and a standard deviation of 0.4 inch.

​(a) What is the probability that a randomly selected board cut by the machine has a length greater than 102.11 ​inches?

​(b) A sample of 43 boards is randomly selected. What is the probability that their mean length is greater than 102.11 inches?

Question 3:

The weights of ice cream cartons are normally distributed with a mean weight of 9 ounces and a standard deviation of 0.4 ounce.

​(a) What is the probability that a randomly selected carton has a weight greater than 9.13 ounces?

​(b) A sample of 36 cartons is randomly selected. What is the probability that their mean weight is greater than 9.13 ounces?

Answer 1

1) z-score = (
エー
)/($\sigma/\sqrt n$)

                 = (555 - 511)/(119/$\sqrt {50}$)

                 = 2.61

Yes, because the z-score is unusual, since it does not lie within 2 standard deviations.

2) P(X > 102.11)

= P((X - $\mu$)/$\sigma$ > (102.11 - $\mu$)/$\sigma$)

= P(Z > (102.11 - 102)/0.4)

= P(Z > 0.28)

= 1 - P(Z < 0.28)

= 1 - 0.6103

= 0.3897

b) P($\bar x$ > 102.11)

= P(($\bar x$ - $\mu$)/($\sigma/\sqrt n$) > (102.11 - $\mu$)/($\sigma/\sqrt n$))

= P(Z > (102.11 - 102)/(0.4/$\sqrt {43}$))

= P(Z > 1.80)

= 1 - P(Z < 1.80)

= 1 - 0.9641

= 0.0359

3)a) P(X > 9.13)

= P((X - $\mu$)/$\sigma$ > (9.13 - $\mu$)/$\sigma$)

= P(Z > (9.13 - 9)/0.4)

= P(Z > 0.33)

= 1 - P(Z < 0.33)

= 1 - 0.6293

= 0.3707

b) P($\bar x$ > 9.13)

= P(($\bar x$ - $\mu$)/($\sigma/\sqrt n$) > (9.13 - $\mu$)/($\sigma/\sqrt n$))

= P(Z > (9.13 - 9)/(0.4/$\sqrt {36}$))

= P(Z > 1.95)

= 1 - P(Z < 1.95)

= 1 - 0.9744

= 0.0256