Question 1

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. 
A) 0.4656
B) 0.0344
C) -0.0344
D) 0.9656

Accepted Answer

The answer of Find the area of the shaded region....

Question 2

An unbiased estimator is a statistic that targets the value of the of the population parameter such that the sampling distribution of the statistic has a ________ equal to the ________ of the corresponding parameter.
A) range; range/4
B) mean; standard deviation
C) mean; mean
D) standard deviation; standard deviation

Accepted Answer

An unbiased estimator is a statistic that on average equals the population parameter it is estimating, hence the sampling distribution of the estimator has an expected value (i.e. mean) equal to the value of the population parameter being estimated. Therefore, the correct answer is C - mean, mean.

Question 3

A baseball player has a batting average of 0.346, so the probability of a hit is 0.346. Assume that his hitting attempts are independent of each other. Assume that the batter gets up to bat 4 times in each game. Estimate the
Probability that in 50 consecutive games, there are at least 45 games in which the batter gets at least one hit.
(Hint: first find the probability that in one game the batter gets at least one hit)
A) 0.0918
B) 0.0446
C) 0.8171
D) 0.0643

Accepted Answer

The answer of A baseball player has a batting average...

Question 4

Scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7. Find $P _ { 81 }$ , which separates the bottom 81% from the top 19%.
A) 0.291
B) 66.6
C) 73.5
D) 0.88

Accepted Answer

We can use the inverse normal distribution function on a calculator to find the z-score associated with the 81st percentile: 
invNorm(0.81) = 0.8779 
Now we can use the z-score formula to find the corresponding raw score: 
z = (x - μ) / σ 
0.8779 = (x - 63.2) / 11.7 
x = 73.5 
Therefore, the score that separates the bottom 81% from the top 19% is 73.5. The answer is choice C.

Question 5

In one region, the September energy consumption levels for single -family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh. For a randomly selected home, find
The probability that the September energy consumption level is between 1100 kWh and 1225 kWh.
A) 0.2881
B) 0.0910
C) 0.3791
D) 0.1971

Accepted Answer

The answer of In one region, the September energy consumption...

Question 6

A coin is tossed 20 times. A person who claims to have extrasensory perception is asked to predict the outcome of each flip in advance. She predicts correctly on 14 tosses. What is the probability of being correct 14 or more
Times by guessing? Does this probability seem to verify her claim? Use the normal distribution to approximate
The desired probability.
A) 0.0582, no
B) 0.4418, no
C) 0.4418, yes
D) 0.0582, yes

Accepted Answer

The probability of correctly guessing the outcome of a coin toss is 0.5. Using the binomial distribution with n = 20 and p = 0.5, the probability of getting 14 or more correct guesses is:

P(X ≥ 14) = 1 - P(X < 14)
= 1 - ∑(from k=0 to k=13) (20 choose k) (0.5)^20
= 1 - 0.9418
= 0.0582

This probability is quite low, which means that the person's claim of having extrasensory perception is not supported by the evidence. To use the normal distribution to approximate the probability, we can assume that the distribution of the number of correct guesses is approximately normal with mean np = 20(0.5) = 10 and standard deviation sqrt(np(1-p)) = sqrt(5) = 2.236. Then, we can use the standard normal distribution to find:

z = (14 - 10) / 2.236 = 1.79

Using a standard normal table or calculator, we find that the probability of getting a z-score of 1.79 or higher is approximately 0.0379. This is close to, but still lower than, the exact binomial probability of 0.0582. Therefore, the best answer is A) 0.0582, no.

Question 7

If z is a standard normal variable, find the probability: $P ( - 0.73 < z < 2.27 )$
A) 0.4884
B) 0.7557
C) 1.54
D) 0.2211

Accepted Answer

Using a standard normal distribution table or a calculator, we find that the area to the left of -0.73 is 0.2317 and the area to the left of 2.27 is 0.9884. Therefore, the area between -0.73 and 2.27 is 0.9884 - 0.2317 = 0.7567. Rounded to four decimal places, the answer is 0.7557, which corresponds to choice B.

Question 8

The probability that a radish seed will germinate is 0.7. Estimate the probability that of 140 randomly selected seeds, exactly 100 will germinate.
A) 0.0679
B) 0.9331
C) 0.0769
D) 0.0669

Accepted Answer

The problem can be solved using the normal approximation to the binomial distribution. The mean (μ) is np = 140 * 0.7 = 98, and the standard deviation (σ) is sqrt(np(1-p)) = sqrt(140 * 0.7 * 0.3) ≈ 5.42. Using the normal approximation, the probability of exactly 100 seeds germinating is approximately 0.0669.

Question 9

The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds. If a college football player is randomly selected, find the probability that he weighs
Between 170 and 220 pounds.
A) 0.1554
B) 0.3811
C) 0.0703
D) 0.2257

Accepted Answer

To find the probability that the player weighs between 170 and 220 pounds, we need to find the area under the normal curve between those two values. We can standardize the distribution by using the z-score formula:

z = (x - μ) / σ

Where x is the weight we are interested in, μ is the mean weight, and σ is the standard deviation.

For x = 170:

z = (170 - 200) / 50 = -0.6

For x = 220:

z = (220 - 200) / 50 = 0.4

Now we can use a standard normal distribution table (or a calculator that has this function) to find the area under the curve between these two z-scores. The answer is:

P(-0.6 < z < 0.4) = 0.3811

Therefore, the best choice is B.

Question 10

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200
And 275.
A) 0.5
B) 0.9332
C) 0.0668
D) 0.4332

Accepted Answer

The probability of a rating between 200 and 275 is found by calculating the area under the normal distribution curve between these two values. The Z-score for 275 is (275-200)/50 = 1.5. The probability of a Z-score less than 1.5 is 0.9332. Since the mean is 200, the probability of a rating less than 200 is 0.5. Therefore, the probability of a rating between 200 and 275 is 0.9332 - 0.5 = 0.4332.

Question 11

The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability.
The probability of no more than 35 defective CDs.
A) The area to the left of 35.5
B) The area to the right of 35.5
C) The area to the left of 35
D) The area to the left of 34.5

Accepted Answer

The continuity correction involves adding 0.5 to the discrete value when using a normal distribution to approximate a discrete distribution. For "no more than 35," we consider the area to the left of 35.5.

Question 12

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that
Of 10,000 randomly selected hair dryers, at least 219 are defective.
A) 0.0934
B) 0.0823
C) 0.0869
D) 0.9066

Accepted Answer

Using the normal approximation to the binomial distribution, we find the mean (μ = np = 10,000 * 0.02 = 200) and standard deviation (σ = √(np(1-p)) = √(10,000 * 0.02 * 0.98) ≈ 14). We then calculate the z-score for 218.5 (continuity correction) and find the probability that z is greater than this value, which is approximately 0.0934.

Question 13

Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Approximately what percentage of
Women have red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter?
A) 4.09%
B) 82.31%
C) 16.11%
D) 17.69%

Accepted Answer

To find the percentage of women with red blood cell counts in the normal range from 4.2 to 5.4 million cells per microliter, we need to find the z-scores for both values and use the standard normal distribution table.

Z-score for 4.2: (4.2 - 4.577) / 0.382 = -0.981
Z-score for 5.4: (5.4 - 4.577) / 0.382 = 2.160

Using the standard normal distribution table, we find that the percentage of women with red blood cell counts between these two z-scores is approximately 82.31%. Therefore, the best choice is B.

Question 14

Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Find the value closest to the probability
That a randomly selected woman has a red blood cell count above the normal range of 4.2 to 5.4 million cells per
Microliter.
A) 0.1611
B) 0.9842
C) 0.0409
D) 0.0158

Accepted Answer

The answer of Assume that the red blood cell counts...

Question 15

The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0°C at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing
Point of water, some give readings below 0°C (denoted by negative numbers) and some give readings above
0°C (denoted by positive numbers). Assume that the mean reading is 0°C and the standard deviation of the
Readings is 1.00°C. Also assume that the frequency distribution of errors closely resembles the normal
Distribution. A thermometer is randomly selected and tested. A quality control analyst wants to examine
Thermometers that give readings in the bottom 4%. Find the temperature reading that separates the bottom 4%
From the others.
A) -1.48°
B) -1.75°
C) -1.89°
D) -1.63°

Accepted Answer

The bottom 4% of a normal distribution corresponds to a z-score of approximately -1.75. Given the mean is 0°C and the standard deviation is 1°C, the temperature reading that separates the bottom 4% is -1.75°C.

Question 16

If z is a standard normal variable, find P(z > 0.97).
A) 0.1685
B) 0.8340
C) 0.1660
D) 0.1922

Accepted Answer

Using a standard normal distribution table, we find the area to the right of 0.97 is 0.1660. Therefore, the probability of z being greater than 0.97 is 0.1660.

Question 17

In a population of 210 women, the heights of the women are normally distributed with a mean of 64.4 inches and a standard deviation of 2.9 inches. If 36 women are selected at random, find the mean $\mu _ { x } ^ { - }$ and standard
Deviation $\sigma _ { x } ^ { - }$ of the population of sample means. Assume that the sampling is done without replacement and
Use a finite population correction factor.
A) 58.8 inches, 2.65 inches
B) 64.4 inches, 2.07 inches
C) 64.4 inches, 2.9 inches
D) 64.4 inches, 0.44 inches

Accepted Answer

The answer of In a population of 210 women, the...

Question 18

Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Find the 80 $80 ^ { \text {th } }$ percentile for the red
Blood cell counts of women.
A) 4.898 million cells per microliter
B) 4.655 million cells per microliter
C) 4.878 million cells per microliter

Accepted Answer

The 80th percentile is the value below which 80% of the data falls. Using a standard normal distribution table, the z-score corresponding to the 80th percentile is 0.84. The 80th percentile for the red blood cell counts of women is then:4.577 + 0.84 * 0.382 = 4.898 million cells per microliter

Question 19

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Shaded area is 0.4483. 
A) 0.3264
B) 0.13
C) 0.6736
D) -0.13

Accepted Answer

The answer of Find the indicated z score. The graph...

Question 20

Which of the following is a biased estimator?
A) mean
B) variance
C) proportion
D) standard deviation

Accepted Answer

The answer of Which of the following is a biased...

Quiz 6: Normal Probability Distributions

Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

An unbiased estimator is a statistic that targets the value of the of the population parameter such that the sampling distribution of the statistic has a equal to the of the corresponding parameter.

Scores on a test are normally distributed with a mean of 63.2 and a standard deviation of 11.7. Find $P _ { 81 }$ , which separates the bottom 81% from the top 19%.

If z is a standard normal variable, find the probability: $P ( - 0.73 < z < 2.27 )$

The probability that a radish seed will germinate is 0.7. Estimate the probability that of 140 randomly selected seeds, exactly 100 will germinate.

The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds. If a college football player is randomly selected, find the probability that he weighs Between 170 and 220 pounds.

A bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected, find the probability of a rating that is between 200 And 275.

The given values are discrete. Use the continuity correction and describe the region of the normal distribution that corresponds to the indicated probability. The probability of no more than 35 defective CDs.

Estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution. Two percent of hair dryers produced in a certain plant are defective. Estimate the probability that Of 10,000 randomly selected hair dryers, at least 219 are defective.

If z is a standard normal variable, find P(z > 0.97) .

Assume that the red blood cell counts of women are normally distributed with a mean of 4.577 million cells per microliter and a standard deviation of 0.382 million cells per microliter. Find the 80 $80 ^ { \text {th } }$ percentile for the red Blood cell counts of women.

Find the indicated z score. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Shaded area is 0.4483.

Which of the following is a biased estimator?