13.) Seventy three percent of drivers carry jumper cables in their vehicles. For a random sample of 315 drivers, Binomial Distributions  (Show how you get these either by formula or calculator) a.) What is the probability that exactly 215 carry jumper cables in their vehicle?b.) What is the probability that at most 220 carry jumper cables in their vehicles?c.) What is the probability that more than 212 carry jumper cables in their vehicles?d.) What is the mean value of the number of drivers that carry jumper cables in their vehicle?e.) What is the standard deviation?

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Asked Jun 12, 2019
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13.) Seventy three percent of drivers carry jumper cables in their vehicles. For a random sample of 315 drivers, Binomial Distributions  (Show how you get these either by formula or calculator)

 

a.) What is the probability that exactly 215 carry jumper cables in their vehicle?

b.) What is the probability that at most 220 carry jumper cables in their vehicles?

c.) What is the probability that more than 212 carry jumper cables in their vehicles?

d.) What is the mean value of the number of drivers that carry jumper cables in their vehicle?

e.) What is the standard deviation?

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Expert Answer

Step 1

The probability that a driver carry jumper cables in the vehicle (p) is 0.73.

Sample size (n) = 315

Let X be the number of drivers in the sample who carry jumper cables in their vehicles, which follows binomial distribution with n=315 and p=0.73.

As n is large, the required probabilities can be calculated using Excel function “BINOMDIST”.

Step 2

a)

The required probability is P(X = 215). The calculation is shown below:

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-BINOM.DIST(215,315,0.73, FALSE) E G 0.008468 Thus, P(X 215) - |0.0085 b) The required probability is P(X 220) fx BINOM.DIST(220,315,0.73,TRUE) E E 0.115952 Thus, P(X 220 = 0.116

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Step 3

c)

The required probability is P(...

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f 1-BINOM.DIST(212,315,0.73,TRUE) E D E G 0.985402 Thus, P(X 212) = 0.9854|

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