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

Question 2

Explain how a nonstandard normal distribution differs from the standard normal
distribution. Describe the process for finding probabilities for nonstandard normal
distributions.

Accepted Answer

The answer of Explain how a nonstandard normal distribution differs...

Question 3

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

The answer of Scores on a test are normally distributed...

Question 4

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 5

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 answer of Estimate the indicated probability by using the...

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

Accepted Answer

The probability of guessing correctly on a coin toss is 0.5. Therefore, the probability of guessing correctly 14 or more times out of 20 can be found using the binomial distribution: P(X ≥ 14) = 1 - P(X < 14) where X is the number of correct guesses and P(X < 14) is the cumulative probability of getting less than 14 correct guesses. Using a binomial calculator, we find: P(X < 14) = 0.9463 Therefore: P(X ≥ 14) = 1 - 0.9463 = 0.0537 This probability is less than 0.05, which is typically considered statistically significant. Therefore, we can say that her claim is not verified by this result. We can also use the normal approximation to the binomial distribution to find the same probability. For the normal approximation, we need to calculate the mean and standard deviation of the distribution of correct guesses: mu = n * p = 20 * 0.5 = 10 sigma = sqrt(n * p * (1 - p)) = sqrt(20 * 0.5 * 0.5) = 2.236 We can then use the normal distribution to approximate the desired probability: P(X ≥ 14) ≈ P(Z > (14 - 10) / 2.236) = P(Z > 1.79) Using a normal probability table or calculator, we find: P(Z > 1.79) = 0.0367 This is the same as the probability we found using the binomial distribution. Therefore, the best choice is C: 0.0582, no.

Question 7

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 consumption level is between 1100 kWh and 1225 kWh, we need to standardize the values and use the standard normal distribution table.

First, we standardize 1100 kWh:

z = (1100 - 1050) / 218 = 0.2294

Next, we standardize 1225 kWh:

z = (1225 - 1050) / 218 = 0.8046

Using a standard normal distribution table, we find the area to the left of 0.2294 is 0.5905, and the area to the left of 0.8046 is 0.7915. Therefore, the probability that the consumption level is between 1100 kWh and 1225 kWh is:

0.7915 - 0.5905 = 0.201

Rounding to four decimal places, we get 0.1971, which is closest to option C).

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. 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

This problem requires finding the percentage of women with red blood cell count between 4.2 and 5.4 million cells per microliter. To solve this, we need to standardize the values using the formula z = (x - μ) / σ, where x is the value being standardized, μ is the mean, and σ is the standard deviation.

For the lower end of the range, z = (4.2 - 4.577) / 0.382 = -0.982. For the upper end of the range, z = (5.4 - 4.577) / 0.382 = 2.165.

Using a standard normal distribution table or calculator, we can find the area under the curve between these two z-scores, which gives us the percentage of women with red blood cell counts in the normal range:

P(-0.982 < z < 2.165) = P(z < 2.165) - P(z < -0.982) = 0.9846 - 0.1684 = 0.8162 or 81.62%.

Therefore, the answer is A) 82.31% (rounded to nearest percent).

Question 9

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...

Question 10

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 11

Find the area of the shaded region. The graph depicts the standard normal distribution wit? 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 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. Find the 80th 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 answer of Assume that the red blood cell counts...

Question 13

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 14

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 a statistic that targets the value of the population parameter such that the sampling distribution of the statistic has a mean equal to the corresponding parameter. This means that on average, the estimator will not overestimate or underestimate the parameter. Option A correctly identifies that the sampling distribution of the unbiased estimator has a mean equal to the population parameter. Options B and C are incorrect because the standard deviation is not necessarily equal to the parameter. Option D is also incorrect because the range is not a measure of central tendency and has no relationship to the parameter or the sampling distribution.

Question 15

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 16

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

The answer of If $z$ is a standard normal variable,...

Question 17

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

We need to find the probability that the weight of the player is between 170 and 220 pounds.

First, we need to standardize these values using the formula:

z = (x - μ) / σ

where: 
x = weight of the player 
μ = mean weight = 200 pounds 
σ = standard deviation = 50 pounds

For x = 170 pounds: 
z1 = (170 - 200) / 50 = -0.6

For x = 220 pounds: 
z2 = (220 - 200) / 50 = 0.4

Now, we need to find the area under the standard normal distribution curve between z1 and z2:

P(-0.6 < z < 0.4)

Using a standard normal distribution table or calculator, we can find this probability to be approximately 0.3811. Therefore, the answer is B.

Question 18

A normal quartile plot is given below for a sample of scores on an aptitude test. Use the
plot to assess the normality of scores on this test. Explain your reasoning.

Accepted Answer

The answer of A normal quartile plot is given below...

Question 19

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 answer of The Precision Scientific Instrument Company manufactures thermometers...

Question 20

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

Accepted Answer

The answer of \(\text { If } z \text {...

Quiz 6: Normal Probability Distributions

Explain how a nonstandard normal distribution differs from the standard normal distribution. Describe the process for finding probabilities for nonstandard normal distributions.

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 $19 \%$ .

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

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

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 80th percentile for the red blood cell counts of women.

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.

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

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

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 normal quartile plot is given below for a sample of scores on an aptitude test. Use the plot to assess the normality of scores on this test. Explain your reasoning.

$\text { If } z \text { is a standard normal variable, find } P ( z > 0.97 ) \text {. }$