702 0 10 wer questions 16-17. blem III. In a recent survey of 100 people, 52 ind

Question

702 0 10 wer questions 16-17. blem III. In a recent survey of 100 people, 52 indicated that they had flown on an airplane at least once, and s indicated that they had ridden on a train at least once. Only 10 of the respondents indicated that they had Joth flown on an airplane at least once and ridden on a train at least once. 16. What is the probability that a person randomly selected from the sample indicated they had flown on an airplane at least once but had never ridden on a train? a) 52% b) 42% c) 18% d) 10% 17. What is the probability that a person randomly selected from the sample indicated they had ridden on a train at least once but had never flown on an airplane? a) 52% b) 28% c) 18% d) 10% PART II. (Use Problem IV to answer questions 18 and 19) Problem IV. Automobiles leaving the paint department of an assembly plant are subjected to a detailed examination of all exterior painted surfaces. For the most recent 380 automobiles produced, the number of blemishes per car is summarized in the follow table. Blemishes per car Number of cars 242 94 38 18. What is the average or mean of this distribution, that is, the average number of blemishes per car? a) .5 b) 2 c) 38 d) 76 19. What is the standard deviation of this distribution? a) .223 b) .414 c) .500 d) 707 20. What is the probability that an automobile will have at least 1 blemishes or P(x21)? a) .3033 b) .3935 c) .6065 d) .9098 Problem V. Of the 51,163 residents of Harrisburg, Arkansas, 20 percent were born outside of Arkansas. A group of 5 is to be randomly selected from this city, and the discrete random variable is x = the number of persons in the group who were born outside of Arkansas. 21. What is the probability that at most 4 or P(x

Explanation / Answer

Q21.

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 5 * 0.2
= 1
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 5 * 0.2 * 0.8
= 0.8
III.
standard deviation = sqrt( variance ) = sqrt(0.8)
=0.894427
P( X < = 4) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 5 4 ) * 0.2^4 * ( 1- 0.2 ) ^1 + ( 5 3 ) * 0.2^3 * ( 1- 0.2 ) ^2 + ( 5 2 ) * 0.2^2 * ( 1- 0.2 ) ^3 +
( 5 1 ) * 0.2^1 * ( 1- 0.2 ) ^4 + ( 5 0 ) * 0.2^0 * ( 1- 0.2 ) ^5
= 0.99968

[ANSWER] d. 99.97%

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