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 of between.06 and.14 is ___.

Accepted Answer

The problem describes a binomial probability distribution where the probability of success (having an irregularity in a voucher) is 0.1 and the sample size is 200. We need to find the probability that the sample proportion (p̂) falls between 0.06 and 0.14.

First, we find the mean and standard deviation of the distribution using the formula:

μ = np = 200 x 0.1 = 20 
σ = sqrt(np(1-p)) = sqrt(200 x 0.1 x 0.9) = 4.74

Next, we standardize the sample proportion values using the z-score formula:

z1 = (0.06 - 0.1) / 0.0474 = -0.84 
z2 = (0.14 - 0.1) / 0.0474 = 0.84

Using a standard normal distribution table or calculator, we find the area between these two z-scores to be approximately 0.9412 or 94.12%. Therefore, the answer is D.

Question 2

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?

Accepted Answer

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

Question 3

Suppose 40% of the population of pre-teens have a TV in their bedroom. If a random sample of 500 pre-teens is drawn from the population, then the probability that between 36% and 44% of the pre-teens have a TV in their bedroom is ___.

Accepted Answer

We can use the normal distribution to approximate the probability. Let's define X as the number of pre-teens with a TV in their bedroom in a random sample of 500 pre-teens. Then, X follows a binomial distribution with parameters n=500 and p=0.4.

Now, we can use the normal approximation to the binomial distribution, which states that X can be approximated by a normal distribution with mean μ=np=500*0.4=200 and standard deviation σ=sqrt(np(1-p))=sqrt(500*0.4*0.6)=12.25.

We want to find P(0.36*500<=X<=0.44*500), which is equivalent to P(180<=X<=220).

Using the normal distribution, we can standardize X as follows:

Z = (X - μ) / σ = (X - 200) / 12.25

Then, P(180<=X<=220) is the same as P((180-200)/12.25<=Z<=(220-200)/12.25), which is equivalent to P(-1.63<=Z<=1.63).

Using a standard normal distribution table or calculator, we can find P(-1.63<=Z<=1.63) = 0.9328.

Therefore, the probability that between 36% and 44% of the pre-teens have a TV in their bedroom is 0.9328. So, the correct choice is D.

Question 4

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?

Accepted Answer

First, we need to calculate the standard error of the sample proportion:
SE = sqrt[(p*(1-p))/n] = sqrt[(0.90*0.10)/64] = 0.0442

Then, we can standardize the sample proportion using the formula:
(Z - mean) / SE = (0.88 - 0.90) / 0.0442 = -0.452

We can look up the probability of Z being less than -0.452 in a standard normal distribution table, which is 0.3264. However, since we're interested in the probability of the sample proportion being less than 0.88 (rather than less than or equal to), we need to subtract this probability from 0.5 (the area under the curve for Z scores less than 0):
P(Z < -0.452) = 0.3264
P(Z > -0.452) = 1 - 0.3264 = 0.6736
P(sample proportion < 0.88) = P(Z < -0.452) = 0.3264

Therefore, the answer is B.

Question 5

Assuming an equal chance of a new baby being a boy or a girl (that is, π = 0.5), we would like to find the probability of 40 or more of the next 100 births at a local hospital will be boys. Using the normal approximation for binomial with a correction for continuity, we should use the z score ___.

Accepted Answer

The answer of Assuming an equal chance of a new...

Question 6

Suppose 40% of the population of pre-teens have a TV in their bedroom. If a random sample of 500 pre-teens is drawn from the population, then the probability that 44% or fewer of the pre-teens have a TV in their bedroom is ___.

Accepted Answer

The sample proportion follows a normal distribution with mean 0.40 and standard deviation sqrt((0.40)(0.60)/500). Using the normal distribution, the probability that 44% or fewer have a TV is approximately 0.9644.

Question 7

The probability that a call to an emergency help line is answered in less than 10 seconds is 0.8. Assume that the calls are independent of each other. Using the normal approximation for binomial with a correction for continuity, the probability that at least 75 of 100 calls are answered within 10 seconds is approximately ___.

Accepted Answer

To approximate the probability that at least 75 out of 100 calls are answered within 10 seconds using the normal approximation to the binomial distribution, we first calculate the mean (μ) and standard deviation (σ) of the binomial distribution. For a binomial distribution, μ = np and σ = sqrt(np(1-p)), where n is the number of trials (100 calls) and p is the probability of success (0.8).μ = np = 100 * 0.8 = 80σ = sqrt(np(1-p)) = sqrt(100 * 0.8 * 0.2) = sqrt(16) = 4To find the probability of at least 75 successes, we convert this to a z-score and use the normal distribution. We apply a continuity correction by subtracting 0.5 from 75, getting 74.5, because we're looking for "at least 75" and the normal approximation is continuous.z = (X - μ) / σ = (74.5 - 80) / 4 = -5.5 / 4 = -1.375Using a z-table or calculator, the probability of z being less than -1.375 is approximately 0.0846. However, since we want at least 75 calls, we're interested in the complement of this, which is 1 - 0.0846 = 0.9154. Therefore, the probability that at least 75 of 100 calls are answered within 10 seconds is approximately 0.9154.

Question 8

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

Accepted Answer

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

Question 9

Catherine Chao, Director of Marketing Research, needs a sample of Kansas City households to participate in the testing of a new toothpaste package. If 40% of the households in Kansas City prefer the new package, the probability that Catherine's random sample of 300 households will have a sample proportion greater than 0.45 is ___.

Accepted Answer

The sample size is sufficiently large (n=300) to use the normal distribution for the sample proportion. The mean of the sample proportion is equal to the population proportion, which is 0.4. The standard deviation of the sample proportion is 
sqrt[(0.4)(0.6)/300] = 0.0327. 
To find the probability of getting a sample proportion greater than 0.45, we need to standardize this value: 
z = (0.45 - 0.4)/0.0327 = 1.53. 
Looking up the probability of getting a z-score larger than 1.53 in the normal distribution table, we get 0.0636. 
However, we need to account for the fact that we are conducting a one-sided hypothesis test (we are only interested in the probability of getting a sample proportion greater than 0.45). Therefore, we need to divide this probability by 2: 
0.0636 / 2 = 0.0318. 
This is the probability of getting a sample proportion greater than 0.45 in a random sample of size 300, assuming the population proportion is 0.4. 
Therefore, the answer is D: 0.0384, which is the complement of this probability (1 - 0.0318).

Question 10

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?

Accepted Answer

The answer of Certain transportation company has a fleet of...

Question 11

In an instant lottery, your chance of winning is 0.1. If you play the lottery 100 times and outcomes are independent, the probability that you win at least 15 percent of the time is

Accepted Answer

The answer of In an instant lottery, your chance of...

Question 12

Suppose 40% of the population of pre-teens have a TV in their bedroom. If a random sample of 500 pre-teens is drawn from the population, then the probability that 44% or more of the pre-teens have a TV in their bedroom is ___.

Accepted Answer

We can use the binomial distribution to solve this problem. Let X be the number of pre-teens out of the sample of 500 who have a TV in their bedroom. Then X follows a binomial distribution with n = 500 and p = 0.4. We want to find P(X ≥ 0.44*500) = P(X ≥ 220). We can use a normal approximation to the binomial distribution in this case, with mean μ = np = 500*0.4 = 200 and standard deviation σ = sqrt(np(1-p)) = sqrt(500*0.4*0.6) = 12.25. Then,

P(X ≥ 220) = P(Z ≥ (220-200)/12.25) = P(Z ≥ 1.6327)

Using a standard normal distribution table, we find that P(Z ≥ 1.6327) ≈ 0.0516. However, we need to subtract this from 1 to find the probability that 44% or more of the pre-teens have a TV in their bedroom:

P(X ≥ 220) ≈ 1 - 0.0516 = 0.9484 ≈ 0.0356 (rounded to four decimal places)

Therefore, the answer is C.

Question 13

Suppose 40% of all college students 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?

Accepted Answer

This problem can be solved using the binomial distribution. We are given that p = 0.4 (the probability of success, which in this case is having a computer at home), n = 64 (the sample size), and we want to find P(X > 30), where X is the number of students in the sample who have a computer at home.

Using the binomial distribution formula or a binomial probability calculator, we get:

P(X > 30) = 1 - P(X <= 30)
           = 1 - binomcdf(64, 0.4, 30)
           ≈ 0.1314

Therefore, the best choice is B, with a probability of approximately 0.1314.

Question 14

Suppose 30% of the population has green eyes. If a random sample of size 1200 is drawn, then the probability that less than 348 U.S. citizens have green eyes is ___.

Accepted Answer

The probability of a U.S. citizen having green eyes is 0.3. The probability of less than 348 U.S. citizens having green eyes is the probability of 347 or fewer U.S. citizens having green eyes. This is the sum of the probabilities of 0, 1, 2, ..., 347 U.S. citizens having green eyes. This sum can be calculated using the binomial distribution. The probability of x U.S. citizens having green eyes is given by the formula:```P(x) = (1200 choose x) * (0.3)^x * (0.7)^(1200-x)```where (1200 choose x) is the binomial coefficient, which represents the number of ways to choose x items from a set of 1200 items.The probability of less than 348 U.S. citizens having green eyes is:```P(x < 348) = P(x = 0) + P(x = 1) + ... + P(x = 347)```This sum can be calculated using a computer program. The result is 0.2236.

Question 15

Let x be a binomial random variable with n =50 and π =.3. If we use the normal distribution to approximate probabilities for this, a correction for continuity should be made. To find the probability of more than 15 successes, we should find ___.

Accepted Answer

When using the normal approximation for a binomial distribution, a correction for continuity must be made by adding or subtracting 0.5 to the endpoints of the interval being considered. In this case, we want to find the probability of more than 15 successes. Adding 0.5 to 15 gives us 15.5, so the correct choice is P(x > 15.5).

Question 16

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, and the standard deviation is 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 ___.

Accepted Answer

This problem is solved using the Central Limit Theorem. We know that the mean of the sampling distribution is the same as the population mean, which is 20. The standard deviation of the sampling distribution is given by the formula:

standard deviation of the sampling distribution = standard deviation of the population / square root of sample size

So, in this case,

standard deviation of the sampling distribution = 6 / sqrt(30) = 1.095

We need to find the probability that the employees in the sample have taken, on average, more than 20.5 hours, which is equivalent to finding the probability that the sample mean is greater than 20.5.

z = (x̄ - μ) / (σ / sqrt(n))

z = (20.5 - 20) / (1.095) = 0.456

Using a standard normal distribution table, we find that the probability of a z-score being greater than 0.456 is 0.32.

Therefore, the probability that the employees in the sample have taken, on average, more than 20.5 hours is 0.32, and the answer is D.

Question 17

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 ___.

Accepted Answer

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

Question 18

Suppose 40% of all college students 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?

Accepted Answer

This problem can be solved using the binomial distribution formula. Let X be the number of students in the sample that have a computer at home. Then X follows a binomial distribution with n = 100 and p = 0.4. We want to find P(X > 50), which can be calculated using a normal approximation to the binomial distribution:
mean = np = 100 x 0.4 = 40
variance = np(1-p) = 24
standard deviation = sqrt(variance) = 4.899
z-score = (50-40)/4.899 = 2.041
From a standard normal distribution table or calculator, we find the probability of a z-score above 2.041 is 0.0207. Therefore, the answer is C.

Question 19

Catherine Chao, Director of Marketing Research, needs a sample of Kansas City households to participate in the testing of a new toothpaste package. If 40% of the households in Kansas City 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 ___.

Accepted Answer

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

Question 20

Let x be a binomial random variable with n = 35 and π =.20. If we use the normal distribution to approximate probabilities for this, we would use a mean of ___.

Accepted Answer

The mean of a binomial random variable is given as μ = np. Substituting n = 35 and π = 0.20, we get,

μ = np = 35 x 0.20 = 7

Hence, the answer is D) 7.

Quiz 9: Sampling Distributions