1.Suppose X is a normal random variable with mean = 100 and standard deviation =

Question

1.Suppose X is a normal random variable with mean = 100 and standard deviation = 5. Find b such that P(100 X b) = 0.3. HINT [See Example 3.] (Round your answer to one decimal place.)

2.Suppose the following are the city driving gas mileages of a selection of sport utility vehicles (SUVs).

19, 20, 19, 20, 18, 21, 17, 19, 24, 23, 21, 21, 17, 20, 20, 18

(a) Find the sample standard deviation (rounded to two decimal places).


(b) In what gas mileage range does Chebyshev's inequality predict that at least 75% of the selection will fall? (Enter your answer using interval notation. Round your answers to two decimal places.)

( , )

(c) What is the actual percentage of SUV models of the sample that fall in the range predicted in part (b)? (Round your answer to two decimal places.)
%
3.According to a study, the probability that a randomly selected teenager shopped at a mall at least once during a week was 0.65. Let X be the number of students in a randomly selected group of 40 that will shop at a mall during the next week.

(a) Compute the expected value and standard deviation of X. (Round your answer to two decimal places.) Hint [See Example 5.]

(b) Fill in the missing quantity. (Round your answer to the nearest whole number.)

There is an approximately 2.5% chance that  or more teenagers in the group will shop at the mall during the next week.

Explanation / Answer

1) MEAN = 100

STANDARD DEVIATION = 5

THE DISTRIBUTION IS NORMAL THEREFORE FORMULA TO BE USED

Z = (X-MEAN)/STANDARD DEVIATION

P(100<X<B) = 0.3

For x = 100 , z = (100 - 100) / 5 = 0 and for

x = B, z = (B - 100) / 5

Hence P(100 < x < B) = P(0 < z < (B-100)/5) = [area to the left of z = (B-100)/5] - [area to the left of 0]

THE PROBABILITY = 0.3

THEREFORE Z SCORE = 1.88

HENCE

(B-100)/5 - 0 = 1.88

THEREFORE B = 1.88*5+100

= 109.4

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