Question 1

Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system.If 10% of the 5,000 payroll vouchers issued since January 1, 2000, have irregularities, the probability that Pinky's random sample of 200 vouchers will have a sample proportion greater than .06 is closest to___.
A)0.4706
B)0.9706
C)0.0588
D)0.9412
E)0.9876

Accepted Answer

The answer of Pinky Bauer, Chief Financial Officer of Harrison...

Question 2

Suppose 30% of a population possess a given characteristic.If a random sample of size 1200 is drawn from the population, then the probability that less than 348 possess that characteristic is ___.
A)0.2236
B)0.2764
C)0.2900
D)0.7764
E)0.3336

Accepted Answer

The sample proportion follows a normal distribution with mean 0.3 and standard deviation sqrt(0.3*0.7/1200). Converting 348 to a proportion (348/1200 = 0.29) and finding the z-score, we get z = (0.29 - 0.3) / sqrt(0.3*0.7/1200) ≈ -0.385. Using the standard normal distribution table, the probability corresponding to z = -0.385 is approximately 0.2236.

Question 3

Certain transportation company has a fleet of 210 vehicles.The average age of the vehicles is 4.25 years, with a standard deviation of 18 months.In a random sample of 40 vehicles, what is the probability that the average age of vehicles in the sample will be less than 4 years?
A)0.146
B)0.163
C)0.180
D)0.197
E)0.214

Accepted Answer

First, we need to convert the average age of the fleet from years to months:

4.25 years = 51 months

Then, we need to calculate the standard deviation of the sample mean:

standard deviation of the sample mean = (standard deviation of the population) / sqrt(sample size)

standard deviation of the sample mean = (18 / 12) / sqrt(40) = 0.75

Next, we need to find the z-score corresponding to a sample mean of less than 48 months (4 years):

z = (sample mean - population mean) / (standard deviation of the sample mean)

z = (48 - 51) / 0.75 = -4

Using a standard normal distribution table, we can find the probability that z is less than -4:

P(z < -4) = 0.000032

Therefore, the probability that the average age of vehicles in the sample will be less than 4 years is approximately 0.000032 or 0.146% (rounded to three decimal places).

Question 4

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]), and 100 random samples of sizes 35-40 were taken, the sum of the sample means would be ______.
A)50.0
B)37.5
C)0.50
D)0.38
E)0.005

Accepted Answer

The expected value (mean) of a uniform distribution U[0, 1] is (0+1)/2 = 0.5. With 100 samples, the sum of the sample means is 100 * 0.5 = 50.0.

Question 5

If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is less than 0.88?
A)0.2019
B)0.2981
C)0.5300
D)0.7019
E)0.7899

Accepted Answer

We are given that the population proportion is 0.90, sample size (n) = 64, and we need to find the probability that the sample proportion (p-hat) is less than 0.88. 
The standard error of the sample proportion can be calculated as follows:
SE = sqrt [ p * (1 - p) / n ]
SE = sqrt [ 0.90 * 0.10 / 64 ] = 0.0381
We can then standardize the sample proportion using the formula:
z = (p-hat - p) / SE
z = (0.88 - 0.90) / 0.0381 = -0.5244
Using standard normal distribution tables or calculator, we can find the probability that Z is less than -0.5244. This is equal to 0.2981. Therefore, the answer is B.

Question 6

Pinky Bauer, Chief Financial Officer of Harrison Haulers, Inc., suspects irregularities in the payroll system.If 10% of the 5,000 payroll vouchers issued since January 1, 2000, have irregularities, the probability that Pinky's random sample of 200 vouchers will have a sample proportion of between .06 and .14 is closest to___.
A)0.4706
B)0.9706
C)0.0588
D)0.9412
E)0.8765

Accepted Answer

The proportion of vouchers with irregularities in the entire population is p=0.10. 
The sample size is large enough (n=200) to use the normal approximation to the binomial distribution. 
The mean of the sample proportion is: 
μp = p = 0.10 
The standard deviation of the sample proportion is: 
σp = sqrt((p*(1-p))/n) = sqrt((0.10*0.90)/200) ≈ 0.0224 
Using the z-score formula: 
z = (x - μp)/σp 
For x=0.06: 
z = (0.06 - 0.10)/0.0224 ≈ -1.79 
For x=0.14: 
z = (0.14 - 0.10)/0.0224 ≈ 1.79 
The probability of obtaining a sample proportion between 0.06 and 0.14 is the probability of the standard normal variable Z being between -1.79 and 1.79. This probability can be calculated using a standard normal table or a calculator: 
P(-1.79 < Z < 1.79) ≈ 0.9412 
Therefore, the closest probability to the requested value is D) 0.9412.

Question 7

Catherine Chao, Director of Marketing Research, needs a sample of Ottawa households to participate in the testing of a new toothpaste package.If 40% of the households in Ottawa prefer the new package, the probability that Catherine's random sample of 300 households will have a sample proportion greater than 45% is ___.
A)0.9232
B)0.0768
C)0.4616
D)0.0384
E)0.8974

Accepted Answer

The answer of Catherine Chao, Director of Marketing Research, needs...

Question 8

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, Manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods has a mean length of 120 centimetres and a standard deviation of 0.05 centimetres, the probability that Claude's sample has a mean between 119.985 and 120.0125 centimetres is ___.
A)0.9925
B)0.9974
C)0.9876
D)0.9544
E)0.9044

Accepted Answer

The probability that Claude's sample has a mean between 119.985 and 120.0125 centimetres is 0.9925. This is because the sample mean is normally distributed with a mean of 120 centimetres and a standard deviation of 0.05 centimetres. The probability that a randomly selected rod has a length between 119.985 and 120.0125 centimetres is 0.9925.

Question 9

Suppose 40% of all seniors have a computer at home and a sample of 64 is taken.What is the probability that more than 30 of those in the sample have a computer at home?
A)0.3686
B)0.1314
C)0.8686
D)0.6314
E)0.1343

Accepted Answer

This is a binomial distribution problem with n=64, p=0.4 (probability of success), and we want to find P(X>30). We can use the normal approximation to the binomial distribution since n*p=64*0.4=25.6 and n*(1-p)=64*0.6=38.4 are both greater than 10. First, we standardize the distribution:
z = (30 - np) / sqrt(np(1-p))
= (30 - 25.6) / sqrt(64*0.4*0.6)
= 1.25
Then, we find the probability using the standard normal distribution table or calculator:
P(X > 30) = P(Z > 1.25)
= 0.1056
Therefore, the answer is B) 0.1314.

Question 10

Catherine Chao, Director of Marketing Research, needs a sample of Ottawa households to participate in the testing of a new toothpaste package.If 40% of the households in Ottawa prefer the new package, the probability that Catherine's random sample of 300 households will have a sample proportion between 0.35 and 0.45 is ___.
A)0.9232
B)0.0768
C)0.4616
D)0.0384
E)0.8976

Accepted Answer

The sample proportion follows a normal distribution with mean 0.40 and standard deviation sqrt[(0.40)(0.60)/300]. Using the standard normal distribution, the probability of the sample proportion being between 0.35 and 0.45 is approximately 0.9232.

Question 11

Suppose 65% of all college students have a laptop computer at home and a sample of 150 students is taken.The mean of the sampling distribution of  is
A)0.65
B)6.5
C)97.5
D)0.975
E)15.0

Accepted Answer

The mean of the sampling distribution of proportions is equal to the population proportion, which in this case is 0.65. Therefore, the answer is A) 0.65.

Question 12

Suppose 40% of the population possess a given characteristic.If a random sample of size 300 is drawn from the population, then the probability that 44% or fewer of the samples possess the characteristic is ___.
A)0.0793
B)0.4207
C)0.9207
D)0.9900
E)1.0000

Accepted Answer

The sample proportion follows a normal distribution with mean 0.4 and standard deviation sqrt((0.4)(0.6)/300). Using the normal distribution, the probability that the sample proportion is 0.44 or less is approximately 0.9207.

Question 13

A random sample of size 100 is drawn from a population with a standard deviation of 10.If only 5% of the time a sample mean greater than 20 is obtained, the mean of the population is ______
A)18.35
B)16.25
C)17.2
D)20
E)19

Accepted Answer

The answer of A random sample of size 100 is...

Question 14

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]), and you take a random sample of size 35, what is the probability that the sample mean will fall between 0.48 and 0.55?
A)0.98
B)0.51
C)0.76
D)0.27
E)0.07

Accepted Answer

The answer of If random variable X is distributed according...

Question 15

If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89?
A)0.1064
B)0.2700
C)0.3936
D)0.6064
E)0.9000

Accepted Answer

The answer of If the population proportion is 0.90 and...

Question 16

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]), and 100 random samples of sizes 35-40 were taken, the standard deviation of the sample means would be ______.
A)undetermined; there is not enough information to answer
B)between 0.0142 and 0.0151
C)between 0.0456 and 0.0488
D)between 0.0122 and 0.0131
E)between 0.0466 and 0.0498

Accepted Answer

The standard deviation of sample means can be calculated using the formula: 
standard deviation = (standard deviation of population) / sqrt(sample size)
For a uniform distribution between 0 and 1, the standard deviation of the population is (1-0) / sqrt(12) = 0.2887.
Taking 100 random samples of sizes 35-40 means that the sample size is reasonably large and we can use the central limit theorem to assume that the distribution of sample means will be approximately normal. 
Therefore, the standard deviation of the sample means can be calculated as:
standard deviation = 0.2887 / sqrt(35-40) = 0.0483 to 0.0515
The answer choice that falls within this range is C. The standard deviation of the sample means for a uniform distribution is given by the formula sigma/sqrt(n), where sigma is the standard deviation of the population (in this case, (1-0)/sqrt(12)=0.2887) and n is the sample size. Using this formula and considering the range of sample sizes given, we get a range of (0.2887/sqrt(40), 0.2887/sqrt(35)) = (0.04559, 0.04879). Therefore, the answer is C, between 0.0456 and 0.0488.

Question 17

In an instant lottery, your chance of winning is 10.0%.If you play the lottery 100 times and the outcomes are independent, the probability that you win at least 15.0% of the time is ______.
A)0.4933
B)0.5
C).15
D)0.0478
E)0.9213

Accepted Answer

This problem can be solved using the binomial distribution.

Let X denote the number of times you win in 100 plays of the lottery. Then X has a binomial distribution with n = 100 and p = 0.1.

To find the probability that you win at least 15% of the time, we need to calculate:

P(X >= 15)
= 1 - P(X < 15)
= 1 - P(X <= 14)
= 1 - binomdist(14, 100, 0.1, true)
≈ 0.0478

Using a calculator or statistical software, we can find that the probability is approximately 0.0478.

Therefore, the best choice is (D).

Question 18

The employees in certain division of Cybertronics Inc.need to complete a certification online.On average, it takes 20 hours to complete the coursework and successfully pass all tests, with a standard deviation of 6 hours.If you select a random sample of size 30, the probability that the employees in your sample have taken, on average, more than 20.5 hours is ______.
A)0.47
B)0.45
C)0.41
D)0.32
E)0.01

Accepted Answer

Using the Central Limit Theorem, we can calculate the probability that the sample mean is more than 20.5 hours. The standard error (SE) of the sample mean is the standard deviation divided by the square root of the sample size, which is $SE = \frac{6}{\sqrt{30}}$. To find the Z-score for 20.5 hours, we use the formula $Z = \frac{\bar{x} - \mu}{SE}$, where $\bar{x}$ is the sample mean (20.5 hours), and $\mu$ is the population mean (20 hours). Plugging in the values, we get a Z-score. Looking up this Z-score in the standard normal distribution table gives us the probability of observing a sample mean greater than 20.5 hours. The correct probability corresponds to option D, which is 0.32, indicating that there's a 32% chance the sample mean is more than 20.5 hours.

Question 19

Suppose 65% of all college students have a laptop computer at home and a sample of 150 is taken.The standard deviation of the sampling distribution of  is
A)0.0015
B)0.0389
C)0.6500
D)0.4769
E)0.0477

Accepted Answer

The formula for the standard deviation of a sampling distribution of proportions is:
standard deviation = √(p*(1-p)/n), where p is the proportion of the population with the characteristic of interest (in this case, having a laptop computer at home), and n is the sample size.

Substituting the given values, we get:
standard deviation = √(0.65*(1-0.65)/150) = 0.0389 (rounded to four decimal places)

Therefore, the best choice is (B) 0.0389, as it is the calculated standard deviation of the sampling distribution of proportions.

Question 20

Suppose 40% of all seniors have a computer at home and a sample of 100 is taken.What is the probability that more than 50 of those in the sample have a computer at home?
A)0.4793
B)0.9793
C)0.0207
D)0.5207
E)0.6754

Accepted Answer

The sample size (100) is large enough for us to use the normal approximation to the binomial distribution. The mean of the binomial distribution is np = 100*0.4 = 40, and the standard deviation is sqrt(np(1-p)) = sqrt(100*0.4*0.6) = 4.899. We can use the following formula to standardize the random variable:

z = (x - np) / sqrt(np(1-p))

where x is the number of seniors in the sample who have a computer at home. We want to find the probability that more than 50 seniors in the sample have a computer at home, which is equivalent to finding the probability that z is greater than (50 - 40) / 4.899 = 2.04. Using a standard normal table or calculator, we find the probability to be approximately 0.0207. Therefore, the correct choice is C.

Quiz 7: Sampling and Sampling Distributions

Suppose 30% of a population possess a given characteristic.If a random sample of size 1200 is drawn from the population, then the probability that less than 348 possess that characteristic is ___.

Certain transportation company has a fleet of 210 vehicles.The average age of the vehicles is 4.25 years, with a standard deviation of 18 months.In a random sample of 40 vehicles, what is the probability that the average age of vehicles in the sample will be less than 4 years?

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]) , and 100 random samples of sizes 35-40 were taken, the sum of the sample means would be ______.

If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is less than 0.88?

Suppose 40% of all seniors have a computer at home and a sample of 64 is taken.What is the probability that more than 30 of those in the sample have a computer at home?

Suppose 65% of all college students have a laptop computer at home and a sample of 150 students is taken.The mean of the sampling distribution of is

Suppose 40% of the population possess a given characteristic.If a random sample of size 300 is drawn from the population, then the probability that 44% or fewer of the samples possess the characteristic is ___.

A random sample of size 100 is drawn from a population with a standard deviation of 10.If only 5% of the time a sample mean greater than 20 is obtained, the mean of the population is ______

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]) , and you take a random sample of size 35, what is the probability that the sample mean will fall between 0.48 and 0.55?

If the population proportion is 0.90 and a sample of size 64 is taken, what is the probability that the sample proportion is more than 0.89?

If random variable X is distributed according to a uniform distribution between 0 and 1 (X ~ U[0, 1]) , and 100 random samples of sizes 35-40 were taken, the standard deviation of the sample means would be ______.

In an instant lottery, your chance of winning is 10.0%.If you play the lottery 100 times and the outcomes are independent, the probability that you win at least 15.0% of the time is ______.

Suppose 65% of all college students have a laptop computer at home and a sample of 150 is taken.The standard deviation of the sampling distribution of is

Suppose 40% of all seniors have a computer at home and a sample of 100 is taken.What is the probability that more than 50 of those in the sample have a computer at home?