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Look up the Z-Value 1.87 in the Standard Normal Table. The area associated with this value is ___ (format is decimal and 4 digits, like .1234).(1 pt)

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Yogi Bear was a popular cartoon character in the 1960s. Yogi lived in Jelly Stone National Park, where he liked to outfox Mr. Ranger, and prided himself on being, "Smarter than the average bear."

Lake Wobegon is a fictional town in Minnesota, the mythical home of the popular radio program "A Prairie Home Companion." As narrator Garrison Keller will tell you, Lake Wobegon is a place, "Where all the women are strong, all the men are good looking, and all the children are above average."

What do Yogi Bear and the children of Lake Wobegon have in common? (1 pt)

Select one:

a. Positive intelligence Z-scores

b. Negative intelligence Z-Scores

c. "0" intelligence Z-Scores

d. Cannot be determined from the information given.

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Reference the Procedure Table in Bluman (pages 269-70 of 3rd ed, 287-8 of 4th ed). Draw out the type of problem described below, and simply indicate which type of problem is being described (Paradigm 1 - 7).

Find the area between 0 and a negative Z-value. This is what paradigm? (1 pt)

Select one:

a. 1

b. 2

c. 3

d. 4

e. 5

f. 6

g. 7

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Reference the Procedure Table in Bluman (pages 269-70 of 3rd ed, 287-8 of 4th ed). Draw out the type of problem described below, and simply indicate which type of problem is being described (Paradigm 1 - 7).

Find the area between a negative Z-value and a positive Z-value. This is what paradigm?(1 pt)

Select one:

a. 1

b. 2

c. 3

d. 4

e. 5

f. 6

g. 7

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What is the P(z>1.22)?

(Hint: If you draw out this question correctly, it will appear like the first picture under Paradigm 2 of your text. Use the method described there to solve the problem.) (2 pts)

Select one:

a. 0.1261

b. 0.3849

c. 0.1112

d. 0.3888

e. 0.6222

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What is P(-0.75 < z < 1.0)? (1 pt)

Select one:

a. 0.2734

b. 0.3413

c. 0.3853

d. 0.5000

e. 0.6147

f. 0.8542

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What is the P(z<0)? (1 pt)

Select one:

a. 0

b. 1 (or 100%)

c. 0.5 (or 50%)

d. -0.5

e. Cannot be determined from the information given.

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The mean score on a 100 question true-false exam is 56 with a standard deviation of 4. A monkey taking the exam scores an expected value of 50.

Find the Z-Value corresponding to the monkey's test score. (Use this value in the next step.)

Determine the approximate percentile rank of the monkey?  

(Hint: Asking for percentile rank is the same as asking what percentage of students did the monkey outscore? To solve this question draw out the picture using the class average and the Monkey's Z-Score, identify the corresponding paradigm from your text, and follow the method given there.)   (2 pts)

Select one:

a. Between 1 and 2 percent.

b. Between 4 and 5 percent.

c. Between 6 and 7 percent.

c. Between 6 and 7 percent.

d. Exactly 10 percent.

e. Between 10 and 11 percent.

f. Exactly 90 percent.

g. Between 93 and 94 percent.

h. Between 97 and 98 percent.

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The distribution of the number of customers not paying their newspaper bills on time is approximately normally distributed. If the mean is 7.9 and the standard deviation is 3.4, what is the probability that the number of non-paying customers is between 9.6 and 13.1?

(Hint: Draw out the problem, identify the paradigm from your text pages 287-8, and follow the method of that paradigm to get the correct answer.) (2 pts)

Select one:

a. 0.4370

b. 0.2455

c. 0.6285

d. 0.5630

e. 0.8085

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This question requires you to find probabilities for an individual. The next question requires you to find the probability for a sample of individuals.  They follow the exact format of Example 6-23 in Bluman (page 303 of the 3rd edition or page 323 of 4th edition). Use Part A as a model for this question, and Part B as a model for the next question. (1 pt)

The average money spent on Product X is $500 per person per year. Assume the standard deviation is $100, and that the distribution is approximately normal. Find the probability that a person selected at random spends less than $550 per year.

Select one:

a. .1915 (or 19.15%)

b. .6915 (or 69.15%)

c. .2985 (or 29.85%)

d. .7985 (or 79.85%)

e. .5000 (or 50%)

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This previous question required you to find probabilities for an individual. This question requires you to find the probability for a sample of individuals.  Both problems follow the exact format of Example 6-23 in Bluman ( page 303 of the 3rd edition or page 323 of the 4th edition). Use Part B as a model for this question. (2 pts)

The average money spent on Product X is $500 per person per year. Assume the standard deviation is $100, and that the distribution is approximately normal. Find the probability that a sample of 25 people selected at random spend less than $550 per year.

Select one:

a. .0072 (or .72%)

b. .4938 (or 49.38%)

c. .9938 (or 99.38%)

d. .9999 (99.99%, or essentially 1, perfect probability)

Explanation / Answer

1) z = 1.87

we have to see the value from z table which comes out to be 0.9693

hence area = 0.9693

in this question u will be able to to see the value just on the table.

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