Please answer and explain: .D. Power and Associates says that 70 percent of car

Question

Please answer and explain:

.D. Power and Associates says that 70 percent of car buyers now use the Internet for research and price comparisons (a) Find the probability that in a sample of 6 car buyers, all 6 will use the Internet. (Round your answer to 4 decimal places.) Probability (b) Find the probability that in a sample of 6 car buyers, at least 3 will use the Internet. (Round your answer to 4 decimal places.) Probability (c) Find the probability that in a sample of 6 car buyers, more than 2 will use the Internet. (Round your answer to 4 decimal places.) l'hobability (d) Find the mean and standard deviation of the probability distribution. (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.) Mean Standard deviation

Explanation / Answer

Solution:

Here, we have to use binomial distribution for finding the required probabilities.

P(X=x) = nCx*p^x*q^(n – x)

Part a

We are given

n = 6, p = 0.70, q = 1 – p = 1 – 0.70 = 0.30

We have to find P(X=6)

P(X=x) = nCx*p^x*q^(n – x)

P(X=6) = 6C6*0.70^6*0.30^(6 – 6)

P(X=6) = 1*0.70^6*0.30^0

P(X=6) = 1*0.70^6*1

P(X=6) = 0.117649

Required probability = 0.1176

Part b

We are given

n = 6, p = 0.70, q = 1 – p = 1 – 0.70 = 0.30

We have to find P(X3)

P(X3) = 1 – P(X<3) = 1 – P(X2)

P(X2) = P(X=0) + P(X=1) + P(X=2)

P(X=0) = 6C0*0.70^0*0.30^6 = 0.000729

P(X=1) = 6C1*0.70^1*0.30^5 = 0.010206

P(X=2) = 6C2*0.70^2*0.30^4 = 0.059535

P(X2) = P(X=0) + P(X=1) + P(X=2)

P(X2) = 0.000729 + 0.010206 + 0.059535

P(X2) = 0.07047

P(X3) = 1 – P(X<3) = 1 – P(X2)

P(X3) = 1 – P(X<3) = 1 – 0.07047

P(X3) = 0.92953

Required probability = 0.9295

Part c

We are given

n = 6, p = 0.70, q = 1 – p = 1 – 0.70 = 0.30

We have to find P(X>2)

P(X>2) = 1 – P(X2)

P(X>2) =1 – 0.07047

P(X>2) =0.92953

Required probability = 0.9295

Part d

We are given

n = 6, p = 0.70, q = 1 – p = 1 – 0.70 = 0.30

Mean = n*p = 6*0.70 = 4.2

Mean = 4.20

Standard deviation = sqrt(npq) = sqrt(6*0.70*0.30) = 1.122497

Standard deviation = 1.1225

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