Question 1

The J.O.Supplies Company buys calculators from a Korean supplier.The probability of a defective calculator is 10%. If 10 calculators are selected at random,what is the probability that 3 or more of the calculators will be defective?

Accepted Answer

P(X F1F1F1S1F1F1F10 3)= 1 - P(X F1F1F1S1F1F1F10 2)= 1 - .9298 = .0702

Question 2

The variable "employment status" which can take on one of two values: 1 for "employed" and 0 for "unemployed" is an example of a _____________ random variable.

Accepted Answer

The variable "employment status" can only take on two discrete values - 1 for "employed" and 0 for "unemployed", hence it is an example of a discrete random variable.

Question 3

A discrete variable that can often be used to describe the number of occurrences of an event over a specified interval of time or space is a _____ random variable.

Accepted Answer

The Poisson distribution is commonly used to model the number of occurrences of an event over a specified interval of time or space, making it the best choice for this scenario. The other options (discrete, hypergeometric, continuous) do not specifically address the counting of occurrences over a defined interval.

Question 4

In a study conducted by UCLA,it was found that 25% of college freshmen support increased military spending.If 6 college freshmen are randomly selected,find the probability that: Only 1 support increased military spending

Accepted Answer

The answer of In a study conducted by UCLA,it was...

Question 5

The J.O.Supplies Company buys calculators from a Korean supplier.The probability of a defective calculator is 10%. If 3 calculators are selected at random,what is the probability that one of the calculators will be defective?

Accepted Answer

To find the probability that one of the calculators will be defective, we can use the binomial distribution formula:

P(X = 1) = (nC1) * p^1 * q^(n-1)

where n = 3 (number of calculators selected), p = 0.1 (probability of a defective calculator), q = 0.9 (probability of a non-defective calculator), and nC1 is the number of ways to select 1 defective calculator out of 3.

nC1 = 3C1 = 3

Plugging these values into the formula, we get:

P(X = 1) = 3 * 0.1^1 * 0.9^2 = 0.243

Therefore, the probability that one of the calculators will be defective is 0.243, which is closest to option C (0.2430).

Question 6

The distribution whose mean is equal to its variance is the _________ distribution.

Accepted Answer

The Poisson distribution is the only distribution where the mean and variance are equal. In the Poisson distribution, the mean and variance are both equal to λ, the parameter of the distribution.

Question 7

In a study conducted by UCLA,it was found that 25% of college freshmen support increased military spending.If 6 college freshmen are randomly selected,find the probability that: Exactly 3 support increased military spending

Accepted Answer

The given problem follows the binomial distribution. Let p be the probability of a freshman supporting increased military spending, then p=0.25 and q=1-p=0.75 (probability of not supporting increased military spending).
The probability of getting exactly 3 freshmen supporting increased military spending out of 6 can be calculated using the binomial distribution formula:
P(X=3) = (6C3) * (0.25)^3 * (0.75)^3
= (6!/3!3!) * (0.25)^3 * (0.75)^3
= 20 * 0.015625 * 0.421875
= 0.1318359375
≈ 0.1318
Hence, the best choice is B.

Question 8

A manufacturer tested a sample of semiconductor chips and found that 35 were defective and 190 were good.If additional tests are to be conducted with random samples of 160 semiconductor chips,find the mean for the number of defects in these groups of 160 (rounded to nearest whole number).

Accepted Answer

The probability of getting a defective chip in a single sample is 35/225 (total number of chips). Therefore, the expected number of defects in a sample of 160 chips is 160*(35/225), which is around 25. Therefore, the correct answer is D.

Question 9

In a study conducted for the State Department of Education,30% of the teachers who left teaching did so because they were laid off.Assume that we randomly select 10 teachers who have recently left their profession.Find the probability that exactly 4 of them were laid off.

Accepted Answer

The answer of In a study conducted for the State...

Question 10

The J.O.Supplies Company buys calculators from a Korean supplier.The probability of a defective calculator is 10%. If 100 calculators are selected at random,what is the standard deviation of the number of defectives?

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula √(n * p * (1-p)), where n is the number of trials and p is the probability of success. Here, n = 100 and p = 0.10, so the standard deviation is √(100 * 0.10 * 0.90) = √9 = 3.00.

Question 11

If x is a binomial random variable,then the standard deviation of x is given by

Accepted Answer

The standard deviation of a binomial random variable is given by the square root of npq, where n is the number of trials, p is the probability of success, and q is the probability of failure (q = 1 - p).

Question 12

It is known that 15% of all college freshmen at Ivey University leave after their first year.Calculate the interval that contains 99.73% of the students who remain if we are looking at a random sample of 300 Ivey University freshmen.

Accepted Answer

Using the normal approximation to the binomial distribution, the interval is calculated as 0.85*300 ± 3*sqrt(0.85*0.15*300).

Question 13

The J.O.Supplies Company buys calculators from a Korean supplier.The probability of a defective calculator is 10%. If 100 calculators are selected at random,what is the expected number of defectives?

Accepted Answer

The expected number of defectives can be calculated by multiplying the probability of a defective calculator (0.10) by the number of calculators selected (100).
Expected number of defectives = 0.10 x 100 = 10. Therefore, the best choice is C.

Question 14

A random variable that is defined to be the total number of successes in n trials is a _____ random variable.

Accepted Answer

A binomial random variable is defined as the total number of successes in n trials, where each trial has only two possible outcomes (success or failure) and the trials are independent and have the same probability of success. The other options (Poisson and Hypergeometric) have different characteristics and are not appropriate for this scenario. A continuous random variable would not be appropriate for counting the number of discrete successes in a set number of trials.

Question 15

The requirement that the probability of success remains constant from trial to trial is a property of the _______________ distribution.

Accepted Answer

The requirement that the probability of success remains constant from trial to trial is a property of the binomial distribution. The other options do not have this same property. The Poisson distribution is used to model the number of rare events occurring in a fixed interval of time, whereas the hypergeometric distribution is used when drawing a sample without replacement from a finite population. The continuous distribution is not specific enough to have a fixed probability of success from trial to trial.

Question 16

Historical data shows that the average number of patient arrivals at the intensive care unit of General Hospital is 3 patients every 2 hours.Assume that the patient arrivals are distributed according to Poisson distribution. Determine the probability of 6 patients arriving in a five-hour period.

Accepted Answer

The probability of 6 patients arriving in a five-hour period can be calculated using the Poisson distribution formula:P(X = k) = (e^-λ * λ^k) / k!where:λ is the average number of arrivals per unit timek is the number of arrivalsIn this case, λ = 3 patients every 2 hours, and k = 6 patients. We need to convert λ to the number of arrivals per 5 hours:λ = 3 patients / 2 hours * 5 hours / 1 = 7.5 patients / 5 hoursNow we can calculate the probability:P(X = 6) = (e^-7.5 * 7.5^6) / 6! = 0.136Therefore, the probability of 6 patients arriving in a five-hour period is 0.136, or 13.6%.

Question 17

In a study conducted by UCLA,it was found that 25% of college freshmen support increased military spending.If 6 college freshmen are randomly selected,find the probability that: Fewer than 4 support increased military spending

Accepted Answer

The answer of In a study conducted by UCLA,it was...

Question 18

A company manufactures an appliance,gives a warranty and 95% of its appliances do not require repair before the warranty expires.If an organization buys 10 of these appliances,calculate an interval that contains 95.44% of all the appliances that won't require repair.

Accepted Answer

The problem involves a binomial distribution with n=10 and p=0.95. The mean is np=9.5 and the standard deviation is sqrt(np(1-p))=0.689. Using the normal approximation, the interval for 95.44% (approximately 2 standard deviations) is [9.5-2*0.689, 9.5+2*0.689] = [8.12, 10.88].

Question 19

A multiple-choice test has 30 questions and each one has five possible answers,of which one is correct.If all answers were guesses,find the probability of getting exactly four correct answers.

Accepted Answer

We can use the binomial distribution formula to solve this problem, where n = 30 (number of trials), p = 1/5 (probability of success - getting a correct answer), and x = 4 (number of successes we want to find the probability of).

Using the formula P(x) = (n choose x) * p^x * (1-p)^(n-x), we get:

P(4) = (30 choose 4) * (1/5)^4 * (4/5)^26
= (27,405) * 0.0016 * 0.000008
= 0.0441

Therefore, the probability of getting exactly four correct answers when guessing on a multiple-choice test with 30 questions is 0.0441 or approximately 0.1325 when rounded to four decimal places. So, the best choice is B.

Question 20

For a random variable X,the mean value of the squared deviations of its values from their expected value is called its

Accepted Answer

The described measure is the variance, which is calculated by finding the average of the squared deviations of X from its expected value. It is a measure of how spread out the values of X are around their mean.

Quiz 5: Discrete Random Variables