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COMM 205- Business Communication QUIZ ASSIGNEMENT CH.6-7 with further explained answers and all you need to know Concordia University CA$18.71   Add to cart
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COMM 205- Business Communication QUIZ ASSIGNEMENT CH.6-7 with further explained answers and all you need to know Concordia University
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COMM 205- Business Communication QUIZ ASSIGNEMENT CH.6-7 with further explained answers and all you need to know Concordia University

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  • June 21, 2024
  • 42
  • 2023/2024
  • Exam (elaborations)
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  • comm 205
  • comm 205 business communication

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COMM 205- Business Communication QUIZ
ASSIGNEMENT CH.6-7 with further
explained answers and all you need to
know Concordia University



ASSIGNEMENT CHAPTER 6

1. Sound City sells the ClearTone-400 satellite car radio. For this radio, historical
sales records over the last 100 weeks show 2 weeks with no radios sold, 18
weeks with one radio sold, 18 weeks with two radios sold, 46 weeks with three
radios sold, 9 weeks with four radios sold, and 7 weeks with five radios sold.
Calculate μx, σ 2, and σ , of x, the number of ClearTone-400
x radios
x sold at Sound

City during a week using the estimated probability distribution. (Round your
answers to 2 decimal places.)

Explanation
(a)
μx = 0 × .02 + 1 × .18 + 2 × .18 + 3 × .46 + 4 × .09 + 5 × .07 = 2.63

(b)
σx2 = (0-2.63)2 × .02 + (1-2.63)2 × .18 + (2-2.63)2 × .18 + (3-2.63)2 × .18 + (4-2.63)2 × .
46 + (5-2.63)2 × .09
= .1383 + .4782 + .0714 + .0630 + .1689 + .3932 = 1.31

(c)
σx = SQRT(1.31) = 1.14

2. Of all individual tax returns, 37 percent include errors made by the taxpayer. If
IRS examiners are assigned randomly selected returns in batches of 12, find
the mean and standard deviation for the number of erroneous returns per
batch.
Multiple Choice
μ = 7.56, σ = 2.80
μ = 4.44, σ = 1.67
Correct
μ = 2.80, σ = 1.67
μ = 4.44, σ = 2.80
Explanation
Mean = np = 12(.37) = 4.44

,Standard Deviation = √(npq) = √[(12)(.37)(.63)] = 1.67

3. The probability distribution of X is

X P(X)
3 1/8
4 1/8
5 3/8
6 3/8

,What is the expected value of X?
Multiple Choice
5.0
Correct
2.24
1.0
2.25
Explanation
E[X]=3(1/8)+4(1/8)+5(3/8)+6(3/8)=40/8=5.0

4. If p = .1 and n = 5, then the corresponding binomial distribution is
.
Multiple Choice
symmetric
right skewed
Correct
left skewed
bimodal
Explanation
Most of the values for this binomial distribution are on the left-hand side of the graph.

5. Using the following probability distribution table of the random
variable x, what is the probability of x = 3?

X P(X)
0 5/15
1 4/15
2 1/15
3

Multiple Choice
5/15
Correct
1/15
3/15
2/15
Explanation
All values of P(X) need to sum to 1, so 5/15 + 4/15 + 1/15 = 10/15 means P(X = 3) = 5/15.

6. Which of the following is a valid probability value for a discrete random
variable?
Multiple Choice
.2

, Correct
• All of the choices are correct.
• −.7
• 1.01
Explanation
The probability of a discrete random variable can only be between 0 and +1.

7. A pharmaceutical company has determined that if a new cholesterol-
reducing drug is manufactured and introduced to the market, the following
probability distribution will describe the contribution of this drug to their
profits during the next six months.

Probability of Profit Contribution
Profit Contribution
−$3,000,000 0.20
(negative profit)
$5,000,000 0.50
$2,000,0000.30


The company management has decided to market this product if the expected
contribution to profit for the next six months is more than
$1,000,000. Based on the information given above, should the company begin
manufacturing the new drug? Explain your answer.
Multiple Choice
• No, do not begin manufacturing.

• Yes, begin manufacturing.Correct
Explanation
μx = .2(−$3,000,000) + .5($5,000,000) + .3($2,000,000) = $2,500,000
$2,500,000 > $1,000,000

A multiple-choice test has 30 questions and each one has five possible answers,
8.
of which only one is correct. If all answers were guesses, find the probability of
getting exactly four correct answers.
Multiple Choice
.1325
Correct
.1227
.0374
.0604
Explanation
Probability of getting question correct = .2. Four successes in 30 trials. Treat as a binomial.

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