Question 1

A small business just leased a new presentation equipment and a color laser printer for three years. The service contract for the computer offers unlimited repairs for a fee of
$100 a year plus a $25 service charge for each repair needed. The company's research
Indicates that during a given year 86% of these computers need no repairs, 9% need to be
Repaired once, 4% twice, 1% three times, and none required more than three repairs.
The standard deviation of the company's annual expense with the service contract for the
Presentation equipment is
A) $105.
B) $10.25.
C) $187.42.
D) $13.69.
E) $2.75.

Accepted Answer

To find the standard deviation of the company's annual expense with the service contract, we first calculate the expected value (mean) and then the variance, from which the standard deviation is the square root.Given probabilities and repair costs:- 86% chance of 0 repairs: $100 annual fee only.- 9% chance of 1 repair: $100 + $25 = $125.- 4% chance of 2 repairs: $100 + 2($25) = $150.- 1% chance of 3 repairs: $100 + 3($25) = $175.Expected value (mean) calculation:$$E(X) = (0.86 \times 100) + (0.09 \times 125) + (0.04 \times 150) + (0.01 \times 175) = 86 + 11.25 + 6 + 1.75 = 105$$Variance calculation:$$Var(X) = (0.86 \times (100-105)^2) + (0.09 \times (125-105)^2) + (0.04 \times (150-105)^2) + (0.01 \times (175-105)^2)$$$$Var(X) = (0.86 \times 25) + (0.09 \times 400) + (0.04 \times 2025) + (0.01 \times 4900)$$$$Var(X) = 21.5 + 36 + 81 + 49 = 187.5$$Standard deviation is the square root of variance:$$SD(X) = \sqrt{187.5} \approx 13.69$$Therefore, the correct answer is D) $13.69.

Question 2

A local unemployment office keeps track of the number of new claims filed each day. Based on data collected, it determines that the expected number of new claims filed per
Day is 2.4 with a standard deviation of 0.8688. Suppose that the office is open five days
Per week. The expected number of new claims filed per week at this office is
A) 7.4.
B) 4.344.
C) 12.
D) 10.
E) 5.25.

Accepted Answer

The expected number of new claims filed per week is calculated by multiplying the daily expected number (2.4) by the number of days the office is open per week (5). Therefore, 2.4 * 5 = 12.

Question 3

For quality control purposes, a company that manufactures sim chips for cell/smart phones routinely takes samples from its production process. Since it is important that
These chips are nearly fault free, one inspection check involves using microscopic
Equipment to count the number of imperfections on each chip. Suppose the average
Number of imperfections per 1000 sim chips is 3.What is the probability that a sample of
This size has no more than 2 imperfections?
A) 0.4232
B) 0.2510
C) 0.2240
D) 0.5689
E) 0.0034

Accepted Answer

The number of imperfections follows a Poisson distribution with λ = 3. The probability of having no more than 2 imperfections is the sum of probabilities of having 0, 1, or 2 imperfections, which is approximately 0.4232.

Question 4

A specific automotive part that a service station stocks in its inventory has an 8% chance of being defective. Suppose many cars come into the service station needing this
Part each week. What is the probability that the tenth part retrieved from stock is the
First defective?
A) 0.2348
B) 0.4344
C) 0.0378
D) 0.4722
E) 0.3780

Accepted Answer

Since the probability of any given part being defective is 0.08, the probability of the 10th part not being defective is (1 - 0.08)^9 ≈ 0.318. Thus, the probability of the 10th part being defective is 1 - 0.318 = 0.682. However, the problem asks for the probability that the 10th part is the first defective, so we need to take into account the fact that the first nine parts may also have been defective. The probability of the first nine parts not being defective is (1 - 0.08)^9 ≈ 0.318, and the probability of the 10th part being defective is still 0.08. Therefore, the probability that the 10th part is the first defective is (0.318)(0.08) = 0.02544, or approximately 0.0378 when rounded to four decimal places.

Question 5

For quality control purposes, a company that manufactures sim chips for cell/smart phones routinely takes samples from its production process. Since it is important that
These chips are nearly fault free, one inspection check involves using microscopic
Equipment to count the number of imperfections on each chip. Suppose the average
Number of imperfections per 1000 sim chips is 3. What is the probability that a sample
This size (1000 chips) has 2 imperfections?
A) 0.4232
B) 0.2510
C) 0.2240
D) 0.5689
E) 0.0034

Accepted Answer

This problem can be solved using the Poisson distribution formula, which is suitable for calculating the probability of a given number of events happening in a fixed interval of time or space, when these events happen with a known constant mean rate and independently of the time since the last event. The formula is $P(k; \lambda) = \frac{\lambda^k e^{-\lambda}}{k!}$, where $P(k; \lambda)$ is the probability of observing $k$ events when the average number of events is $\lambda$, and $e$ is the base of the natural logarithm (approximately 2.71828). Given $\lambda = 3$ (the average number of imperfections per 1000 chips) and $k = 2$ (the number of imperfections we're interested in), the calculation is $P(2; 3) = \frac{3^2 e^{-3}}{2!} = \frac{9 \cdot e^{-3}}{2} \approx 0.2510$.

Question 6

Internet service providers (ISP) need to resolve customer problems as quickly as possible. For one ISP, past data indicates that the likelihood is .80 that customer calls
Regarding Internet service interruptions are resolved within one hour. Out of the next 10
Customer calls about interrupted service, the probability that exactly 7 will be resolved
Within one hour is
A) 0.2013.
B) 0.8791.
C) 0.7897.
D) 0.3452.
E) 0.1209.

Accepted Answer

This is a binomial probability problem where n = 10 (number of trials), k = 7 (number of successes), and p = 0.80 (probability of success on a single trial). The formula for binomial probability is P(X = k) = (n choose k) * p^k * (1-p)^(n-k). Plugging in the values: P(X = 7) = (10 choose 7) * 0.80^7 * 0.20^3 = 0.2013.

Question 7

Suppose that incoming calls per hour to a customer service center of a small credit union are uniformly distributed between 0 and 6 calls. The probability that at least 3 calls
Are received per hour is
A) 3/6.
B) 4/7.
C) 3/7.
D) 4/6.
E) 1/6.

Accepted Answer

The answer of Suppose that incoming calls per hour to...

Question 8

Internet service providers (ISP) need to resolve customer problems as quickly as possible. For one ISP, past data indicates that the likelihood is .80 that customer calls
Regarding Internet service interruptions are resolved within one hour. Out of the next 10
Customer calls about interrupted service, how many would be expected to have their
Service problems resolved within one hour?
A) 2
B) 4
C) 6
D) 8
E) 10

Accepted Answer

The expected number of calls resolved within one hour is calculated by multiplying the total number of calls (10) by the probability of a call being resolved within one hour (0.80), which equals 8.

Quiz 6: Random Variables and Probability Models

A specific automotive part that a service station stocks in its inventory has an 8% chance of being defective. Suppose many cars come into the service station needing this Part each week. What is the probability that the tenth part retrieved from stock is the First defective?

Suppose that incoming calls per hour to a customer service center of a small credit union are uniformly distributed between 0 and 6 calls. The probability that at least 3 calls Are received per hour is