Question

In: Statistics and Probability

Based on a survey, assume that 26% of consumers are comfortable having drones deliver their purchases....

Based on a survey, assume that 26% of consumers are comfortable having drones deliver their purchases. Suppose that we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones. Identify the values of n, x, p, and q.

Solutions

Expert Solution

26% of consumers are comfortable having drones deliver their purchases.

Thus p = 0.26    { probability of success }

And q = 1 - p = 1-0.26 = 0.74

To find the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones.

So here we need to find P(X=3)

Thus here n = 5             { number of random samples/consumers selected }

And here x = 3              {i.e to find exactly three consumers comfortable with delivery by drones }

Thus

n = 5

x = 3

p = 0.26

q = 0.74

Here X ~ B( n = 5 , p = 0.26 )

i.e X follows binomial distribution with probability of success p = 0.26.

Now pmf is given by

P(X=x)   = nCx px q (n-x)

Here we need to find P(X=3)

P(X=3) =  5C3 p3 q (5-3)

P(X=3) = 5C3 * (0.26)3 * (0.74) (5-3)

P(X=3) = 10 * 0.017576 * 0.5476

P(X=3) = 0.09624618

P(X=3) = 0.09625

Thus , the probability that when five consumers are randomly selected, exactly three of them are comfortable with delivery by drones is equal to 0.09625 .


Related Solutions

Based on a survey assume that 35% of consumers are comfortable having drones deliver their purchases....
Based on a survey assume that 35% of consumers are comfortable having drones deliver their purchases. suppose we want to find the probability that when five consumers are randomly selected, exactly three of them are comfortable with the drones. What is wrong with using the multiplication rule to find the probability of three consumers comfortable with drones followed by two consumers not comfortable. A. There are other arrangements consisting of three consumers who are comfortable and two who are not....
A survey by Walmart found that people spend on average $26 on each walk-in shopping. Assume...
A survey by Walmart found that people spend on average $26 on each walk-in shopping. Assume the standard deviation is $4. a) Find the z-score of spending $31.8 ? b) Find P(X > 27) using the first method (Z-score table). Interpret the result. c) Find P(X > 27) using the second method (TI 83/84 calculator). Show all work by drawing the interval on the Normal Distribution curve and indicating the lower and upper values. d) Find the percentage of people...
Based on a smartphone​ survey, assume that 41% of adults with smartphones use them in theaters....
Based on a smartphone​ survey, assume that 41% of adults with smartphones use them in theaters. In a separate survey of 209 adults with​ smartphones, it is found that 71 use them in theaters. a. If the 41​% rate is​ correct, find the probability of getting 71 or fewer smartphone owners who use them in theaters. b. Is the result of 71 significantly​ low?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT