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USDA chicken inspection. In Exercise 3.19 (p. 170), you learned that one in every 100 slaughtered chickens passes USDA inspection with fecal contamination. Consider a random sample of three slaughtered chickens that all pass USDA inspection. Let x equal the number of chickens in the sample that has fecal contamination.

  1. Find p(x)for x = 0, 1, 2, 3.
  2. Graph p(x).
  3. Find P(xโ‰ค1).

Short Answer

Expert verified

a.

b. The graph is given below:

c.12

Step by step solution

01

(a) List of possible outcomes

Let C and N be denoted as contaminated and uncontaminated chickens, respectively, and accordingly, a list of outcomes of 3 randomly picked slaughtered chickens are shown below

02

Calculation of p(x)

The calculations p(x) for each of the values of x representing the number of contaminated slaughtered chickens are shown below.

Therefore, for x equal to 0 and 1, the probability is 18, and for x equal to 1 and 2, it is 38.

03

(b) List of possible outcomes

Here the values of x denote the number of contaminated chickens being randomly picked, and accordingly, the probability distribution table is shown below:

04

Elucidation on the graph

In the graph, the vertical axis consists of the probability distribution values, and the horizontal axis consists of the number of contaminated chickens as represented by x.The blue bars represent the probability distributions of the number of contaminated chickens.

05

(c) List of possible outcomes

In this case, for simplicity, C and N have been chosen to denote contaminated and uncontaminated chickens, respectively, as shown below

06

Calculation of P(xโ‰ค1)

P(xโ‰ค1)means that the three chickens chosen will contain a maximum of one contaminated chicken, and it can be observed that only four types of combinations can occur.

Therefore, by dividing 4 by 8, it P(xโ‰ค1)comes out to be12.

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Most popular questions from this chapter

If x is a binomial random variable, compute for each of the following cases:

  1. n = 4, x = 2, p = .2
  2. n = 3, x = 0, p = .7
  3. n = 5, x = 3, p = .1
  4. n = 3, x = 1, p = .9
  5. n = 3, x = 1, p = .3
  6. n = 4, x = 2, p = .6

Suppose x is a normally distributed random variable with ฮผ= 11 and ฯƒ= 2. Find each of the following:

a)P(10โ‰คฯ‡โ‰ค12)

b) P(6โ‰คฯ‡โ‰ค10)

c)P(13โ‰คฯ‡โ‰ค16)

d)P(7.8โ‰คฯ‡โ‰ค12.6)

e)P(ฯ‡โ‰ฅ13.24)

f)P(ฯ‡โ‰ฅ7.62)


Compute the following:

a.7!3!(7โˆ’3)!

b.(94)

c. (50)

d.(44)

e.(54)


Suppose ฯ‡ is a normally distributed random variable with ฮผ=50 and ฯƒ=3 . Find a value of the random variable, call it ฯ‡0 , such that

a)P(ฯ‡โ‰คฯ‡0)=.8413

b)P(ฯ‡>ฯ‡0)=.025

c)P(ฯ‡>ฯ‡0)=.95

d)P(41โ‰คฯ‡โ‰คฯ‡0)=.8630

e) 10% of the values of role="math" localid="1652160513072" ฯ‡are less thanrole="math" localid="1652160519976" ฯ‡0

f)1% of the values ofฯ‡ are greater thanฯ‡0

Shopping vehicle and judgment. Refer to the Journal of Marketing Research (December 2011) study of whether you are more likely to choose a vice product (e.g., a candy bar) when your arm is flexed (as when carrying a shopping basket) than when your arm is extended (as when pushing a shopping cart), Exercise 2.85 (p. 112). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumers shopping under each of the two conditions. Recall that the average choice score for consumers with a flexed arm was 59, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 5. Also, assume that both distributions are approximately normally distributed.

a. In the flexed arm condition, what is the probability that a consumer has a choice score of 60 or greater?

b. In the extended arm condition, what is the probability that a consumer has a choice score of 60 or greater?

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