Question

(1 point)

A)Suppose the scores of students on an exam are Normally distributed with a mean of 303 and a standard deviation of 89. Then approximately 99.7% of the exam scores lie between the numbers _____and_____ such that the mean is halfway between these two integers. (You are not to use R for this question.)

B)

A telemarketer calls people and tries to sell them a subscription to a daily newspaper. On 16% of her calls, there is no answer or the line is busy. She sells subscriptions to 10% of the remaining calls. For what proportion of calls does she make a sale?

Proportion = _____

C)

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles per hour, and a standard deviation of 4.78 miles per hour

What percent of passenger vehicles travel slower than 80 miles per hour? (Respond to 2 decimal places, do not include percent symbol)
answer: _____%

The speed limit on this stretch of the I-5 is 70 miles per hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. (Respond to 2 decimal places, do not include percent symbol)
answer: ____ %

Answer 1

A)Suppose the scores of students on an exam are Normally distributed with a mean of 303 and a standard deviation of 89. Then approximately 99.7% of the exam scores lie between the numbers _____and_____ such that the mean is halfway between these two integers. (You are not to use R for this question.)

Solution:

We are given that the variable follows normal distribution.

Mean = 303

SD = 89

For the middle 99.7% of the data, the critical Z values by using z-table are given as below:

Z = -2.967737925 and 2.967737925

Required score = Mean + Z*SD

Lower score = 303 - 2.967737925*89 = 38.87132

Upper score = 303 + 2.967737925*89 = 567.1287

Answer: 39 and 567 approximately

B)

A telemarketer calls people and tries to sell them a subscription to a daily newspaper. On 16% of her calls, there is no answer or the line is busy. She sells subscriptions to 10% of the remaining calls. For what proportion of calls does she make a sale?

Solution:

Total calls = 100%

No answer or line busy calls = 16%

Answered calls = 100 – 16 = 84%

10% of remaining 84% = 0.84*0.10 = 0.084 = 8.4%

Required proportion = 8.4%

C)

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles per hour, and a standard deviation of 4.78 miles per hour

What percent of passenger vehicles travel slower than 80 miles per hour? (Respond to 2 decimal places, do not include percent symbol)

Solution:

We are given that the variable is normally distributed.

Mean = 72.6

SD = 4.78

We have to find P(X<80)

Z = (X – mean) / SD

Z = (80 - 72.6)/4.78

Z = 1.548117

P(Z<1.548117) = P(X<80) = 0.939203 = 93.92%

(by using z-table)

Required percent = 93.92

The speed limit on this stretch of the I-5 is 70 miles per hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5. (Respond to 2 decimal places, do not include percent symbol)

Here, we have to find P(X>70)

P(X>70) = 1 – P(X<70)

Mean = 72.6

SD = 4.78

Z = (X – mean) / SD

Z = (70 - 72.6)/4.78

Z = -0.54393

P(Z<-0.54393) = P(X<70) = 0.293244

(by using z-table)

P(X>70) = 1 – P(X<70)

P(X>70) = 1 – 0.293244

P(X>70) = 0.706756 = 70.68%

Required percentage = 70.68