Question 1

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) 0.0643
D) 0.8171

Accepted Answer

The probability that the batter gets at least one hit in one game is 1 - (1 - 0.346)^4 = 0.8171. The probability that the batter gets at least one hit in 50 consecutive games is (0.8171)^50 = 0.0918.

Question 2

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
B) The area to the left of 35.5
C) The area to the right of 35.5
D) The area to the left of 34.5

Accepted Answer

When using the continuity correction for a discrete variable (like the number of defective CDs), you adjust the boundary by 0.5 to account for the discrete nature. For "no more than 35," you would look at the area to the left of 35.5.

Question 3

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) 0.4418, no
B) 0.4418, yes
C) 0.0582, no
D) 0.0582, yes

Accepted Answer

The probability of guessing correctly 14 or more times out of 20 is approximately 0.0582 using the normal distribution. This low probability suggests that her claim of extrasensory perception might be valid, but it is not definitive proof.

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)  73.5 
C)  0.88 
D)  66.6

Accepted Answer

To find $P_{81}$, which separates the bottom 81% from the top 19%, we use the Z-score formula and the normal distribution properties. The Z-score corresponding to 81% in the standard normal distribution table is approximately 0.88. We then use the formula $X = \mu + Z\sigma$, where $X$ is the score we're looking for, $\mu$ is the mean (63.2), $Z$ is the Z-score (0.88), and $\sigma$ is the standard deviation (11.7). Plugging in the values, we get $X = 63.2 + 0.88 \times 11.7 = 73.5$.

Question 5

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) 0.0669
D) 0.0769

Accepted Answer

This is a binomial probability problem where the number of trials n = 140, the number of successes k = 100, and the probability of success on a single trial p = 0.7. The binomial probability formula is P(X = k) = (nCk) * p^k * (1-p)^(n-k), where nCk is the combination of n items taken k at a time. Calculating this directly might be cumbersome, but using a binomial probability calculator or statistical software with these inputs will give a result close to 0.0669, making option C the correct choice.

Question 6

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) 0.3791
B) 0.2881
C) 0.1971
D) 0.0910

Accepted Answer

To find the probability that the energy consumption is between 1100 kWh and 1225 kWh, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For 1100 kWh, $Z = \frac{1100 - 1050}{218} \approx 0.229$, and for 1225 kWh, $Z = \frac{1225 - 1050}{218} \approx 0.802$. Using the standard normal distribution table, the probabilities corresponding to these Z-scores are approximately 0.5904 and 0.7875, respectively. The probability of being between these two values is the difference, 0.7875 - 0.5904 = 0.1971.

Question 7

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) 0.0409
C) 0.0158
D) 0.9842

Accepted Answer

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

Question 8

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^th percentile for the red blood cell counts of women. A) 4.655 million cells per microliter B) 4.898 million cells per microliter C) 4.565 million cells per microliter D) 4.878 million cells per microliter

Accepted Answer

The 80th percentile corresponds to a z-score of approximately 0.84. Using the formula X = μ + zσ, where μ = 4.577 and σ = 0.382, the calculation gives X ≈ 4.898 million cells per microliter.

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) 0.2257
D) 0.0703

Accepted Answer

To find the probability that a player weighs between 170 and 220 pounds, we use the Z-score formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. For X = 170, Z = (170 - 200) / 50 = -0.6. For X = 220, Z = (220 - 200) / 50 = 0.4. Looking these Z-scores up in a standard normal distribution table or using a calculator, we find the probabilities: P(Z < 0.4) - P(Z < -0.6). This gives us approximately 0.6554 - 0.2743 = 0.3811.

Question 10

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

Accepted Answer

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

Question 11

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) mean; mean
B) standard deviation; standard deviation
C) mean; standard deviation
D) range; range/4

Accepted Answer

An unbiased estimator is defined as a statistic where the mean of its sampling distribution is equal to the true mean of the population parameter it estimates. This means that, on average, the estimator hits the target parameter.

Question 12

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) 82.31%
B) 17.69%
C) 4.09%
D) 16.11%

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 use the properties of the normal distribution. First, we calculate the z-scores for the lower and upper limits of the range:- For 4.2 million cells/μL: $Z = \frac{4.2 - 4.577}{0.382} \approx -0.987$- For 5.4 million cells/μL: $Z = \frac{5.4 - 4.577}{0.382} \approx 2.151$Using standard normal distribution tables or a calculator, we find the probabilities corresponding to these z-scores:- The area to the left of $Z = -0.987$ is approximately 0.1619 (16.19%).- The area to the left of $Z = 2.151$ is approximately 0.9842 (98.42%).To find the percentage of women within this range, we subtract the smaller area from the larger area:$0.9842 - 0.1619 = 0.8223$ or 82.23%, which rounds to 82.31% as per the given options.

Question 13

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) 0.0823
B) 0.9066
C) 0.0869
D) 0.0934

Accepted Answer

The normal approximation to the binomial distribution can be used here. The mean (μ) is np = 10,000 * 0.02 = 200, and the standard deviation (σ) is √(np(1-p)) = √(10,000 * 0.02 * 0.98) ≈ 14. The z-score for 218.5 (continuity correction) is (218.5 - 200) / 14 ≈ 1.321. Using the standard normal distribution table, the probability of z being less than 1.321 is approximately 0.9066. Therefore, the probability of at least 219 being defective is 1 - 0.9066 = 0.0934.

Question 14

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) $0.9656$
C) $0.0344$
D) $- 0.0344$

Accepted Answer

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

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) −°1.75
B) −°1.63
C) −°1.48
D) −°1.89

Accepted Answer

The z-score corresponding to the bottom 4% is -1.75. The temperature reading that separates the bottom 4% from the others is therefore 0°C - 1.75°C = -1.75°C.

Question 16

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) 0.4332
B) 0.9332
C) 0.0668
D) 0.5

Accepted Answer

The probability of a rating between 200 and 275 is the area under the normal distribution curve between z-scores of 0 and 1.5. Using a standard normal distribution table, we find that the area between z-scores of 0 and 1.5 is 0.4332.

Question 17

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) $- 0.13$
B) $0.3264$
C) $0.6736$
D) $0.13$

Accepted Answer

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

Question 18

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

Accepted Answer

To find the probability that $z$ is between -0.73 and 2.27, we look up the corresponding values in the standard normal distribution table or use a calculator. The value for $z = 2.27$ is approximately 0.9883 and for $z = -0.73$ is approximately 0.2327. Subtracting these gives $0.9883 - 0.2327 = 0.7556$, which rounds to 0.7557, matching option A.

Question 19

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

Accepted Answer

The answer of If  z  is a standard...

Question 20

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 $\mu _ { \bar { x } }$ and standard deviation $\sigma _ { \bar { x } }$ of the population of sample means. Assume that the sampling is done without replacement and use a finite population correction factor.
A)  64.4  inches,  2.07  inches
B)  58.8  inches,  2.65  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...

Quiz 6: Normal Probability Distributions

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

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% .

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

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^th percentile for the red blood cell counts of women.

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.

Which of the following is a biased estimator?

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.

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.

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

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.

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

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

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