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Based on data from a car bumper sticker? study, when a car is randomly? selected

ID: 3126393 • Letter: B

Question

Based on data from a car bumper sticker? study, when a car is randomly? selected, the number of bumper stickers and the corresponding probabilities are as shown below.

Complete parts? (a) through? (d).

a. Does the given information describe a probability? distribution?

Yes

No

b. Assuming that a probability distribution is? described, find its mean and standard deviation.

The mean is .........................

?(Round to the nearest tenth as? needed.)The standard deviation is...................

c. Use the range rule of thumb to identify the range of values for usual numbers of bumper stickers.

The maximum usual value is ................

?(Round to the nearest tenth as? needed.)The minimum usual value is....................

?(Round to the nearest tenth as? needed.)

d. Is it unusual for a car to have more than one bumper? sticker? Explain.

A.

?No, because the probability of more than 1 bumper sticker is 0.113 which is greater than 0.05.

B.

?Yes, because the probabilities for random variable x from 2 to 9 are all less than 0.05.

C.

?No, because the probability of having 1 bumper sticker is 0.089?, which is greater than 0.05.

D.

Not enough information is given. Click to select your answer(s).

Table of numbers of bumper stickers and probabilities P(x) 0.798 0.089 0.037 0.026 0.014 0.011 0.009 0.007 0.006 0.003 4

Explanation / Answer

a)

YES, because the sum of P(x) values is 1.

******************

b)

Consider:

Thus,  
  
E(x) = Expected value = mean = Sum(xP(x)) =    0.53 [ANSWER, MEAN]

Var(x) = E(x^2) - E(x)^2 =    1.9831

sigma(x) = sqrt [Var(x)] =    1.408225834 [ANSWER, STANDARD DEVIATION]

*****************

c)

Max. Usual = u + 2*sigma(x) = 0.53 + 2*1.408225834 = 3.346451668 [ANSWER]

Min. Usual = u - 2*sigma(x) = 0.53 - 2*1.408225834 = -2.286451668 [ANSWER]

*******************

d)

A. No, because the probability of more than 1 bumper sticker is 0.113 which is greater than 0.05. [ANSWER]

x P(x) x P(x) x^2 P(x) 0 0.798 0 0 1 0.089 0.089 0.089 2 0.037 0.074 0.148 3 0.026 0.078 0.234 4 0.014 0.056 0.224 5 0.011 0.055 0.275 6 0.009 0.054 0.324 7 0.007 0.049 0.343 8 0.006 0.048 0.384 9 0.003 0.027 0.243
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