Problem What are the expected value, variance, and standard deviation of the fol

Question

Problem What are the expected value, variance, and standard deviation of the following distribution: alue of Random ariable utcome robability .3 .2 Arrivals at a fast-food custgmers per hour What is the probabili restaurant follow a Poisson distribution with a mean arrival rate of 16 0,0 0 ity that in the next hour there will be exactly 12 arrivals? of 500 and a standard deviatsed by high school seniors follows a normal distribution with ity that a student uses fewer than 400 minutes? What is the probability that a student uses more Problem 4 Through a telephone survey, a low-interest bank credit card is offered to 500 households. D911 y)60)- The responses are as tabled Income $100,00ncome100.000 Accept offer Reject offer 50 250 40 160 Develop a joint probability table and show the marginal probabilities. What is the probability of a household whose income exceeds $60,000 and who rejects the offer? a. b. c. If income is less than or equal $60,00, what is the probability the offer will be d. If the offer is accepted, what is the probability that income exceeds $60,000? accepted?

Explanation / Answer

(1)

For the given distribution values:

Expected value = E(X) = 1*0.4 + 2*0.3 + 3*0.2 + 4*0.1 = 2

Variance = E(X2)-(E(X))2 =

E(X2) = (1^2)*0.4 + (2^2)*0.3 + (3^2)*0.2 + (4^2)*0.1 = 5

So,

Variance = 5-(2^2) = 1

Standard deviation = Variance0.5 = 1

(3)

Data given:

Mean,m = 500

Standard Deviation, SD = 50

When X = 400, calculating the z-score:

z = (X-m)/SE= (400-500)/50 = -2

The corresponding p-value for this z-score is:

p = 0.02275

So the probability that the student uses fewer than 400 minutes is:

P(X<400) = 0.02275

When X = 350, calculating the z-score:

z = (X-m)/SE= (350-500)/50 = -3

The corresponding p-value for this z-score is:

p = 0.00135

Reqd probability = 1-p = 1-0.00135 = 0.99865

So the probability that the student uses more than 350 minutes is:

P(X>350) = 0.99865

Hope this helps !

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