Chapter 5 review answers PDF

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STAT2201 Business Statistics

Chapter 5 Review Solutions Review Questions For Binomial or Poisson distribution you can use either the tables or formula to arrive at your answers: 1. Complete the following table to calculate the mean, variance, and standard deviation of probability distribution “A”. Distribution A Xi

P(Xi)

XiP(Xi)

[Xi- ]

[Xi-]2

([Xi-]2)P(Xi)

0

0.50

0.00

-1.00

1.00

0.50

1

0.20

0.02

0.00

0.00

0.00

2

0.15

0.30

1.00

1.00

0.15

3

0.10

0.30

2.00

4.00

0.40

4

0.05

0.20

3.00

9.00

0.45

∑= = E(X) = 1.00

∑ = 2 = 1.50 Variance=

 = √𝝈𝟐 = 1.2247

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2. At the Express House Delivery Service, providing high-quality service to customers is the top priority of the management. The company guarantees a refund of all charges if a package it is delivering does not arrive at its destination by the specified time. It is known from past data that despite all efforts, 5% of the packages mailed through this company do not arrive at their destinations within the specified time. If a corporation mails 10 packages through Express House Delivery Service on a certain day. a. Find the probability that exactly one of these 10 packages will not arrive at its destination within the specified time. n = 10, p = 0.05 𝑃(𝑥) = 𝑛𝐶𝑥𝑃 𝑥 (1 − 𝑃)𝑛−𝑥 𝑃(𝑋 = 1) = 10𝐶1 × 0.051 × (1 − 0.05)10−1 = 0.3150 b. Find the probability that at least one of these 10 packages will not arrive at its destination within the specified time. n = 10, p = 0.05 Use 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) find 𝑃(𝑋 = 0) = 10𝐶0 × 0.050 × (1 − 0.05)10−0 = 0.5987 then 𝑃(𝑋 ≥ 1) = 1 − 0.5987 = 0.4013 c. Find the probability of 8 packages arriving at their destinations on time. n = 10, p = 0.95 (complement to probability of arriving late) 𝑃(𝑥) = 𝑛𝐶𝑥𝑃 𝑥 (1 − 𝑃)𝑛−𝑥 𝑃(𝑋 = 8) = 10𝐶8 × 0.958 × (1 − 0.95)10−8 = 0.0746 d. How many of these 10 packages would you expect not to arrive at its destination within the specified time? n = 10, p = 0.05 𝑬(𝑿) = 𝝁 = 𝒏𝒑 = 𝟏𝟎 × 𝟎. 𝟎𝟓 = 0.500 e. What is the variance and standard deviation of the number of packages not arriving at their destination within the specified time? n = 10, p = 0.05 𝑽𝒂𝒓(𝑿) = 𝝈𝟐 = 𝒏𝒑(𝟏 − 𝒑) = 𝟏𝟎 × 𝟎. 𝟎𝟓 × 𝟎. 𝟗𝟓 = 0.4750 𝑺𝑫(𝑿) = 𝝈 = √𝝈𝟐 = √𝟎. 𝟒𝟕𝟓𝟎 = 0.6892 3. An office supply company conducted a survey before marketing a new paper shredder designed for home use. In the survey, 80% of the people who used the shredder were satisfied with it. Because of this high acceptance rate, the company decided to market the new shredder. Assume that 80% of all people who will use it will be satisfied. On a certain day, seven customers bought this shredder. a. How many customers can we expect to be satisfied with their purchase? n = 7, p = 0.80 𝑬(𝑿) = 𝝁 = 𝒏𝒑 = 𝟕 × 𝟎. 𝟖𝟎 = 5.6 b. What is the variance and standard deviation of satisfied customers? 𝑽𝒂𝒓(𝑿) = 𝝈𝟐 = 𝒏𝒑(𝟏 − 𝒑) = 𝟕 × 𝟎. 𝟖𝟎 × 𝟎. 𝟐𝟎 = 1.12 𝑺𝑫(𝑿) = 𝝈 = √𝝈𝟐 = √𝟏. 𝟏𝟐𝟎 = 1.0583

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c. Find the probability that less than 2 will be satisfied. Note: show 7 decimal places to avoid a zero probability Use 𝑃(𝑋 < 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) 𝑃(𝑋 = 0) = 7𝐶0 × 0.80 × (1 − 0.8)7−0 = 0.0000128 𝑃(𝑋 = 1) = 7𝐶1 × 0.81 × (1 − 0.8)7−1 = 0.0003584 𝑃(𝑋 < 2) = 𝟎. 𝟎𝟎𝟎𝟎𝟏𝟐𝟖 + 𝟎. 𝟎𝟎𝟎𝟑𝟓𝟖𝟒 = 0.0003712 d. Find the probability that at least 2 will be satisfied. 𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 < 2) = 1 - 0.0003712 = 0.9996288 4. Despite all efforts by the quality control department, the fabric made at Benton Corporation always contains a few defects. A certain type of fabric made at this corporation contains an average of .5 defects per 500 yards. a. Find the probability that there will be exactly 2 defects in a 500-yard piece of fabric. Note: 𝝁 = 𝟎. 𝟓 Use 𝑷(𝒙) =

𝝀𝒙 𝒆−𝝀 𝒙!

𝑷(𝑿 = 𝟐) =

𝟎.𝟓𝟐 × 𝒆−𝟎.𝟓 𝟐!

=

𝟎.𝟐𝟓𝐱𝟎. 𝟔𝟎𝟔𝟓 𝟐

= 0.0758

b. Find the probability there will be at least one defect in a 500-yard piece of fabric. Use 𝑷(𝑿 ≥ 𝟏) = 𝟏 − 𝑷(𝑿 = 𝟎) 𝟎.𝟓𝟎 ×𝒆−𝟎.𝟓

𝟏𝐱𝟎.𝟔𝟎𝟔𝟓

= 0.6065 = 𝑷(𝑿 = 𝟎) = 𝟏 𝟎! 𝑷(𝑿 ≥ 𝟏) = 𝟏 − 𝟎. 𝟔𝟎𝟔𝟓 = 0.3935 c. Find the probability there will be exactly 1 defect in a given 1000-yard piece of this fabric. Note: 𝝁 = 𝟐 × 𝟎. 𝟓 = 𝟏 𝑷(𝑿 = 𝟏) =

𝟏𝟏 𝐱𝒆−𝟏 𝟏!

= 0.3679

5. You are responsible for quality control at a manufacturing plant with volatile access to electricity. Each month you get an average of 4 power surges. Each time you get a power surge, you lose one day of production. a. What is the probability that you will have no power surges next month? Note: 𝝁 = 𝟒 𝝀𝒙 𝒆−𝝀

𝟒𝟎 ×𝒆−𝟒

𝟏𝐱𝟎.𝟎𝟏𝟖𝟑

𝑷(𝑿 = 𝟎) = 𝟎! = Use 𝑷(𝒙) = 𝒙! = 0.0183 𝟏 b. What is the probability of at least 1 power surge next month? Use 𝑷(𝑿 ≥ 𝟏) = 𝟏 − 𝑷(𝑿 = 𝟎) 𝑷(𝑿 ≥ 𝟏) = 𝟏 − 𝟎. 𝟎𝟏𝟖𝟑 = 0.9817 c. What is the expected number of power surges in the next month? 𝑬(𝑿) = 𝝁 = 4 d. What is the standard deviation of the number power surges in the next month? 𝑺𝑫(𝑿) = 𝝈 = √𝝁 = √𝟒 = 2 note also 𝑽𝒂𝒓(𝑿) = 𝝈𝟐 = 𝝁 = 4

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Multiple Choice Questions

6. Which of the following can be represented by a discrete random variable? A. The circumference of a randomly generated circle B. The time of a flight between Chicago and New York C. The number of defective light bulbs in a sample of five D. The average distance achieved in a series of long jumps 7. Which of the following can be represented by a continuous random variable? A. The average temperature in Tampa, Florida, during the month of July B. The number of typos found on a randomly selected page of this test bank C. The number of students who will get financial assistance in a group of 50 randomly selected students D. The number of customers who visit a department store between 10:00 a.m. and 11:00 a.m. on Mondays 8. Consider the following discrete probability distribution. x P(X=x)

-10 0 10 20 0.35 0.10 0.15 0.40

What is the probability that X is 0? P(X=0) A. 0.10 B. 0.35 C. 0.55 D. 0.65 9. Consider the following cumulative distribution function for the discrete random variable X. x P(X=x)

1 2 3 4 0.30 0.44 0.72 1.00

What is the probability that X is less than or equal to 2? A. B. C. D.

0.44 0.30 0.74 0.56

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10. Which of the following statements is most accurate about a binomial random variable? A. It has a bell-shaped distribution. B. It is a continuous random variable. C. It counts the number of successes in a given number of trials. D. It counts the number of successes in a specified time interval or region.

11. It is known that 10% of the watches shipped from a particular factory are defective. What is the probability that none in a random sample of four watches are defective? A. 0.0010 Look up n = 4 p = 0.10 x = 0 in table or B. 0.2916 C. 0.3439 P(x = 0) = (4C0) (0.10)0 (1- 0.10)4-0 = 0.6561 D. 0.6561

12. Cars arrive randomly at a tollbooth at a rate of 20 cars per 10 minutes during rush hour. What is the probability that exactly five cars will arrive over a five-minute interval during rush hour? If 20 cars in 10 min = λ /5, then λ = 10 A. 0.0378 B. 0.0500 Look up λ = 10 and P(x = 5) in table or C. 0.1251 D. 0.5000 105 𝑒 −10 𝑃(𝑥 = 5) = = 0.0378 5!...