Question 1

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn a "B" on the test, a student must be in the top 15% of the class. What should a student score to receive a "B"?
A) 75.18
B) 75.49
C) 81.48
D) 76.90

Accepted Answer

The top 15% of the class corresponds to a z-score of 1.04. Using the formula z = (x - mu) / sigma, we can solve for x:1.04 = (x - 70) / 5x = 75.18

Question 2

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn a "C" on the test, a student must be in the top 25% of the class. What should a student score to receive a "C"?
A) 77.21
B) 73.37
C) 73.96
D) 76.90

Accepted Answer

To find the score that corresponds to the top 25% of the class, we look for the z-score that corresponds to the 75th percentile (since the top 25% means the lowest score in that range is at the 75th percentile of all scores). The z-score for the 75th percentile is approximately 0.674 (from z-tables or a standard normal distribution calculator). Using the formula for converting z-scores to raw scores: X = μ + Zσ, where X is the raw score, μ is the mean, Z is the z-score, and σ is the standard deviation, we get X = 70 + (0.674)(5) = 70 + 3.37 = 73.37.

Question 3

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the mean for this distribution?
A) 450
B) 90
C) 45
D) There is no mean for this type of distribution

Accepted Answer

The mean of a binomial distribution is given by the formula μ = np, where n is the number of trials (in this case, 50 voters) and p is the probability of success (in this case, 0.9 or 90%). Therefore, the mean is 50 * 0.9 = 45.

Question 4

What is the value of the continuity correction factor?
A) 1.00
B) 0.50
C) 100
D) 1.96

Accepted Answer

The continuity correction factor is used when approximating a discrete distribution with a continuous distribution. It involves adding or subtracting 0.5 from the lower and upper endpoints of a range of values being approximated. In this case, the question does not provide enough information to determine a specific range, so it is best to choose the general value of 0.5 as the continuity correction factor.

Question 5

A statistics student receives a grade of 85 on a statistics midterm. If the corresponding z-score equals +1.5 and the standard deviation equals 7, the average grade on this exam is _______.
A) 74.5
B) 75.4
C) 68.5
D) 76.9

Accepted Answer

The z-score formula is $Z = \frac{(X - \mu)}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. Rearranging to solve for $\mu$, we get $\mu = X - Z\sigma$. Plugging in the given values, $\mu = 85 - (1.5 \times 7) = 85 - 10.5 = 74.5$.

Question 6

Determine the z-score associated with an area of 0.4925 from the mean of a normal distribution to that negative z-score.
A) 2.43
B) -2.43
C) -2.4
D) 2.4
E) -2.44

Accepted Answer

Since the area is to the left of the mean, we need to find the negative z-score. Using a standard normal distribution table, we see that the closest area to 0.4925 is 0.4918, which corresponds to a z-score of 2.43. However, since we need a negative z-score, the answer is -2.43. Therefore, the best choice is B.

Question 7

i. A continuity correction compensates for estimating a discrete distribution with a continuous distribution. ii. The binomial can be used to approximate the normal distribution for small sized samples.
iii. The binomial may be used to approximate the normal distribution when the probability, p, remains the same from trial to trial.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (i) and (ii) are correct statements but not (iii).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

- A continuity correction does indeed compensate for the approximation of a discrete distribution (like binomial) with a continuous one (like normal). The binomial distribution can approximate the normal distribution under certain conditions, notably when the sample size is large (not small as stated in (ii)) and the probability of success p is constant across trials, as correctly noted in (iii).

Question 8

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the probability that fewer than 40 of your sample voters are in favour of strike action?
A) 0.2594
B) 0.4952
C) 0.0048
D) 0.0091

Accepted Answer

The probability can be calculated using the binomial distribution formula with n=50, p=0.9, and finding P(X<40). Using a binomial calculator or statistical software, P(X<40) is approximately 0.0048.

Question 9

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the probability that more than 45 of the voters from your sample voted in favour of strike action?
A) 0.50
B) 0.0948
C) 0.5948
D) 0.4052

Accepted Answer

We can use the binomial distribution formula to solve this problem:

P(X > 45) = 1 - P(X ≤ 45)

where X is the number of voters in favor of strike action in a sample of 50 voters. Since the proportion of voters in favor of strike action is 0.9, the probability of a voter being in favor of strike action is p = 0.9 and the probability of a voter not being in favor of strike action is q = 1 - p = 0.1.

Using the binomial distribution formula, we have:

P(X ≤ 45) = Σi=0^45 (50 choose i) * 0.9^i * 0.1^(50-i)

Using a calculator, we get P(X ≤ 45) ≈ 0.5948.

Therefore,

P(X > 45) = 1 - P(X ≤ 45)
            = 1 - 0.5948
            = 0.4052

So the probability that more than 45 of the voters from the sample voted in favour of strike action is 0.4052. Therefore, the best choice is D.

Question 10

For the standard normal distribution, what z-score corresponds with the 85% percentile?
A) 0.30
B) 2.05
C) 0.15
D) 1.04

Accepted Answer

The 85th percentile of the standard normal distribution corresponds to a z-score of approximately 1.04. This means that 85% of the distribution falls below a z-score of 1.04.

Question 11

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the standard deviation for this distribution?
A) 0.02
B) 1.4
C) 2
D) There is no standard deviation for this type of distribution.

Accepted Answer

The standard deviation for a binomial distribution is calculated as follows:
sqrt(n*p*q)
where n is the number of trials (100 orders), p is the probability of success (98% or 0.98), and q is the probability of failure (2% or 0.02).
Therefore, the standard deviation will be sqrt(100*0.98*0.02) = 1.4, which is option B.

Question 12

i. A continuity correction compensates for estimating a discrete distribution with a continuous distribution. ii. The normal probability distribution is generally deemed a good approximation for the binomial probability distribution when np and n(1 -p)are both greater than five.
iii. When a continuity correction factor is used, its value is 1.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (i) and (ii) are correct statements but not (iii).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

(i) is correct because a continuity correction is used when approximating a discrete distribution with a continuous one.
(ii) is correct because the normal distribution approximation is valid when np and n(1-p) are both greater than 5.
(iii) is false because the value of the continuity correction factor can vary depending on the specific approximation being made.

Question 13

Management is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 percent of production based on past experience. Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. If the bonus is paid on the upper 5 percent of production, the bonus will be paid on how many units or more?
A) 6,255
B) 5,120
C) 3,196
D) 4,099

Accepted Answer

To find the number of units for the top 5% in a normal distribution, we use the Z-score formula. The Z-score for the top 5% is approximately 1.645. The formula to calculate the number of units is: $X = \mu + Z\sigma$, where $X$ is the number of units, $\mu$ is the mean (4,000 units), $Z$ is the Z-score (1.645), and $\sigma$ is the standard deviation (60 units). Plugging in the values: $X = 4000 + 1.645(60) = 4000 + 98.7 \approx 4099$. Therefore, the bonus would be paid on 4,099 units or more.

Question 14

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the probability that there will be more than 5 order errors in a given day?
A) 0.0062
B) 0.4838
C) 0.9838
D) 2.1428

Accepted Answer

We need to find the probability of more than 5 order errors in a day given the error rate of 2%. Since this is a binomial distribution problem, we can use the formula for cumulative probability: P(X > 5) = 1 - P(X <= 5) To find P(X <= 5), we can use the binomial probability formula: P(X = k) = (n choose k) * p^k * (1-p)^(n-k) where n is the total number of orders (100), p is the probability of error (0.02), and k is the number of errors (0 to 5). So, P(X <= 5) = sum of [ (100 choose k) * 0.02^k * 0.98^(100-k) ] for k = 0 to 5 This can be calculated using a calculator or spreadsheet, and the result is approximately 0.9938. Therefore, P(X > 5) = 1 - P(X <= 5) = 1 - 0.9938 = 0.0062. Thus, the answer is A.

Question 15

A new drug has been developed that is found to relieve nasal congestion in 90 percent of those with the condition. The new drug is administered to 300 patients with the condition. What is the probability that more than 265 will be relieved of nasal congestion?
A) 0.0916
B) 0.1922
C) 0.8078
D) 0.3078

Accepted Answer

First, we need to check if the sample size is large enough to use a normal approximation. Since np = 300 * 0.9 = 270 >= 10 and n(1-p) = 300 * (1-0.9) = 30 >= 10, we can use the normal approximation to the binomial distribution. 
Next, we need to calculate the mean and standard deviation of the sample, assuming that the distribution of the sample proportion is normal: 
mean = np = 270 
standard deviation = sqrt(np(1-p)) = sqrt(300*0.9*0.1) = 4.74 
Finally, we can use the normal distribution to find the probability that more than 265 patients will be relieved of nasal congestion: 
z = (265.5 - 270) / 4.74 = -0.95 
P(z > -0.95) = 0.8078 (using a standard normal table or calculator) 
Therefore, the probability that more than 265 patients will be relieved of nasal congestion is 0.8078 or approximately 81%.

Question 16

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the mean for this distribution?
A) 0.02
B) 1.4
C) 2
D) There is no mean for this type of distribution.

Accepted Answer

The answer of Bottomline Ink, a forms managementcompany, fills 100...

Question 17

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn an "A" on the test, a student must be in the top 5% of the class. What should a student score to receive an "A"?
A) 80.84
B) 75.49
C) 78.2
D) 76.90

Accepted Answer

To be in the top 5%, the z-score must be equal to or greater than 1.645. Using the formula z = (x - μ) / σ, we can solve for the score necessary to achieve this z-score:
1.645 = (x - 70) / 5
x = 78.2. Therefore, a student must score at least 78.2% on the test to receive an "A."

Question 18

i. A continuity correction compensates for estimating a discrete distribution with a continuous distribution. ii. The normal probability distribution is generally deemed a good approximation for the binomial probability distribution when np and n(1 -p)are both greater than five.
iii. When a continuity correction factor is used, its value is 0.5.
A) (i), (ii), and (iii) are all correct statements.
B) (i) is a correct statement but not (ii) or (iii).
C) (i) and (iii) are correct statements but not (ii).
D) (i) and (ii) are correct statements but not (iii).
E) (i), (ii), and (iii) are all false statements.

Accepted Answer

A continuity correction is used when approximating a discrete distribution with a continuous one, such as using the normal distribution to approximate the binomial distribution. The normal approximation is generally considered appropriate when both np and n(1-p) are greater than five. The continuity correction factor is typically 0.5.

Question 19

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the standard deviation for this distribution?
A) 45
B) 4.5
C) 2.12
D) 20.25

Accepted Answer

The probability of a vote in favor of the strike is 0.9. 
The probability of a vote against the strike is 1 - 0.9 = 0.1. 
The mean, or expected value, of the distribution is: μ = np = 50 x 0.9 = 45. 
The variance of the distribution is: σ^2 = np(1-p) = 50 x 0.9 x 0.1 = 4.5. 
Therefore, the standard deviation is the square root of the variance: σ = sqrt(4.5) ≈ 2.12.

Question 20

Determine the z-score associated with an area of.1950 from the mean of a normal distribution to that positive z-score.
A) 0.41
B) 0.51
C) 0.5
D) 0.01
E) -0.41

Accepted Answer

Using a standard normal table or calculator, we can find that the z-score associated with an area of 0.1950 to the right of the mean is approximately 0.51. Therefore, the answer is B.

Quiz 6: Continuous Probability Distributions

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn a "B" on the test, a student must be in the top 15% of the class. What should a student score to receive a "B"?

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn a "C" on the test, a student must be in the top 25% of the class. What should a student score to receive a "C"?

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the mean for this distribution?

What is the value of the continuity correction factor?

A statistics student receives a grade of 85 on a statistics midterm. If the corresponding z-score equals +1.5 and the standard deviation equals 7, the average grade on this exam is _______.

Determine the z-score associated with an area of 0.4925 from the mean of a normal distribution to that negative z-score.

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the probability that fewer than 40 of your sample voters are in favour of strike action?

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the probability that more than 45 of the voters from your sample voted in favour of strike action?

For the standard normal distribution, what z-score corresponds with the 85% percentile?

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the standard deviation for this distribution?

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the probability that there will be more than 5 order errors in a given day?

A new drug has been developed that is found to relieve nasal congestion in 90 percent of those with the condition. The new drug is administered to 300 patients with the condition. What is the probability that more than 265 will be relieved of nasal congestion?

Bottomline Ink, a forms managementcompany, fills 100 orders a day with a 2% error rate in the completed orders. Assume this to be a binomial distribution. What is the mean for this distribution?

From past history, the scores on a statistics test are normally distributed with a mean of 70% and a standard deviation of 5%. To earn an "A" on the test, a student must be in the top 5% of the class. What should a student score to receive an "A"?

Tabulation of a strike vote showed that 90% of those voting cast their ballot in favour of strike action. You take a sample of 50 voters. Assume this to be a binomial distribution. What is the standard deviation for this distribution?

Determine the z-score associated with an area of.1950 from the mean of a normal distribution to that positive z-score.