Question 1

A researcher is interested in the lengths of Salvelinus fontinalis (brook trout), which are known to be approximately Normally distributed with mean 80 centimeters and standard deviation 5 centimeters. To help preserve brook trout populations, some regulatory standards need to be set limiting the size of fish that can be caught. The standard is set so that all fish smaller than 70.20 centimeters or larger than 89.80 centimeters must be thrown back. What proportion of the population would be thrown back?

Accepted Answer

We need to find the proportion of the population that falls outside the limits of 70.20 centimeters and 89.80 centimeters.

Using the z-score formula: 
z = (x - μ) / σ

For the lower limit: 
z = (70.20 - 80) / 5 = -1.96

For the upper limit: 
z = (89.80 - 80) / 5 = 1.96

Using a standard normal distribution table, the proportion of the population outside these limits is:

P(z < -1.96 or z > 1.96) = 0.05

Therefore, 5% of the population would be thrown back.

Question 2

Serum albumin levels in adults are distributed approximately Normally with a mean of 35 g/l and a standard deviation of 5 g/l.
What is the probability that a randomly selected adult has a serum albumin level greater than 32.5 g/l?

Accepted Answer

The answer of Serum albumin levels in adults are distributed...

Question 3

Serum albumin levels in adults are distributed approximately Normally with a mean of 35 g/l and a standard deviation of 5 g/l.
What serum albumin level do the lower 10% of adults have?

Accepted Answer

The answer of Serum albumin levels in adults are distributed...

Question 4

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg.
What is the approximate probability that a randomly selected man with liver cirrhosis has a total body protein of 9.75 kg or greater?

Accepted Answer

We need to find the probability that a man with liver cirrhosis has a total body protein of 9.75 kg or greater.

First, we need to standardize the cutoff value using the Z-score formula:

Z = (X - μ) / σ

Z = (9.75 - 9.8) / 0.1 = -0.5

Using a standard Normal distribution table or calculator, we can find the probability of a Z-score of -0.5 or greater to be approximately 0.6914.

Therefore, the approximate probability that a randomly selected man with liver cirrhosis has a total body protein of 9.75 kg or greater is 0.6914.

Question 5

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg.
What is the approximate probability that a randomly selected man with liver cirrhosis has a total body protein between 9.75 and 9.85 kg?

Accepted Answer

The answer of The distribution of total body protein in...

Question 6

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg.
What is the minimum total body protein that the upper 25% of adult men with cirrhosis have?

Accepted Answer

The upper 25% corresponds to the 75th percentile, which in a normal distribution is represented by a Z-score of approximately 0.674 (from Z-tables). Using the formula for converting Z-scores to raw scores: X = μ + Zσ, where μ is the mean, σ is the standard deviation, and Z is the Z-score, we get X = 9.8 + (0.674)(0.1) = 9.8 + 0.0674 = 9.8674, which rounds to 9.87 kg.

Question 7

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg.
We obtain the numerical value for P(X ≤ 9.65). Which of the following statements about the probability, P(X < 9.65), is true?

Accepted Answer

In continuous probability distributions, such as the normal distribution, the probability of any single, exact value is zero. Therefore, P(X < 9.65) is the same as P(X ≤ 9.65).

Question 8

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg.
What is the approximate probability that a randomly selected healthy man has a total body protein greater than 12 kg?

Accepted Answer

The answer of The distribution of total body protein in...

Question 9

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg.
What is the approximate probability that a randomly selected healthy man has a total body protein between 12.1 and 12.2 kg?

Accepted Answer

We need to calculate the probability of a randomly selected healthy man having a total body protein between 12.1 and 12.2 kg. 
Using the standard Normal distribution formula, we can standardize the values as follows:
z1 = (12.1 - 12.3) / 0.1 = -2
z2 = (12.2 - 12.3) / 0.1 = -1
Now, we need to find the area under the Normal curve between -2 and -1. 
Using a standard Normal distribution table or calculator, we can find this area to be approximately 0.1359. 
Therefore, the answer is B.

Question 10

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg.
What is the minimum total body protein for the upper 75% of healthy adult men?

Accepted Answer

To find the minimum total body protein for the upper 75%, we need to find the 25th percentile of the distribution. For a standard normal distribution, the z-score corresponding to the 25th percentile is approximately -0.675. Using the formula: X = μ + zσ, where μ = 12.3 kg and σ = 0.1 kg, we get X = 12.3 + (-0.675)(0.1) = 12.2325 kg, which rounds to 12.23 kg.

Question 11

The distribution of vitamin C amount in vitamin drops produced by a given factory is approximately Normal, with a mean of 60.0 mg and a standard deviation of 0.5 mg.
What is the approximate proportion of vitamin drops containing more than 60.75 mg of vitamin C?

Accepted Answer

To find the proportion of vitamin drops containing more than 60.75 mg of vitamin C, we calculate the z-score for 60.75 mg. The z-score is given by $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = \frac{(60.75 - 60.0)}{0.5} = 1.5$. Looking up the z-score of 1.5 in a standard normal distribution table or using a calculator, we find that the area to the left of z = 1.5 is approximately 0.9332, which means that about 93.32% of the distribution is below 60.75 mg. Therefore, the proportion above 60.75 mg is 1 - 0.9332 = 0.0668, or approximately 7%.

Question 12

The distribution of vitamin C amount in vitamin drops produced by a given factory is approximately Normal, with a mean of 60.0 mg and a standard deviation of 0.5 mg.
What is the approximate probability of drawing at random a vitamin drop with a vitamin content between 60.25 and 60.75 mg?

Accepted Answer

The answer of The distribution of vitamin C amount in...

Question 13

The amount of cholesterol in a person's body produced by the liver and other cells is proposed to be Normally distributed with mean 75% and standard deviation 0.5%.
What is the probability that a person produces more than 76.7% of the cholesterol in his or her body?

Accepted Answer

The Z-score for 76.7% is calculated as (76.7 - 75) / 0.5 = 3.4. The probability of a Z-score greater than 3.4 is approximately 0.0003.

Question 14

The amount of cholesterol in a person's body produced by the liver and other cells is proposed to be Normally distributed with mean 75% and standard deviation 0.5%.
What is the first quartile of the distribution of cholesterol production?

Accepted Answer

To find the first quartile, we need to find the value that separates the lowest 25% of the distribution from the rest. We can use the z-score formula to find the corresponding value:

z = (Q1 - μ) / σ

where Q1 is the first quartile, μ is the mean, σ is the standard deviation, and z is the z-score that corresponds to the 25th percentile, which is -0.6745 (from standard normal distribution tables).

Substituting the given values, we get:

-0.6745 = (Q1 - 0.75) / 0.005

Solving for Q1, we get:

Q1 = 0.75 + (-0.6745) x 0.005 = 0.7466 or 74.66%

Therefore, the first quartile of the distribution of cholesterol production is 74.66%. Option A is correct.

Question 15

X has a Normal distribution with mean 50 and standard deviation 8. Which of the following statements is correct?

Accepted Answer

In a continuous distribution like the Normal distribution, the probability that X is less than or equal to a certain value (P(X ≤ 52)) is the same as the probability that X is strictly less than that value (P(X < 52)). Therefore, P(X < 52) = P(X ≤ 52) in this case.

Question 16

Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 micron.
What is the probability that a binding site position is between 2.0 and 3.0 microns, approximately?

Accepted Answer

To find the probability that a binding site position is between 2.0 and 3.0 microns, we need to find the area under the Normal distribution curve between those two values. We can use a standard Normal distribution and standardize the values by subtracting the mean and dividing by the standard deviation:

Z1 = (2.0 - 2.45) / 0.35 = -1.29
Z2 = (3.0 - 2.45) / 0.35 = 1.57

Using a standard Normal distribution table or calculator, we can find the probability of the Z-score being between -1.29 and 1.57 to be approximately 0.843. Therefore, the answer is B.

Question 17

Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 micron.
What position (in microns) are 20% of binding sites greater than?

Accepted Answer

To find the position that 20% of binding sites are greater than, we need to find the z-score at the 20th percentile using a standard Normal distribution table or calculator.

z-score = invNorm(0.2) = -0.84

Then, we use the formula:

x = z*SD + mean

x = -0.84*0.35 + 2.45

x = 2.73

Therefore, 20% of binding sites are greater than approximately 2.73 microns. The closest choice is C, 2.74 microns, which is the best answer.

Question 18

Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 microns.
What is the standardized score for a binding site position of 2.03 microns?

Accepted Answer

The answer of Using fluorescent imaging techniques, researchers observed that...

Question 19

A researcher was interested in the lengths of brook trout and obtained the following quantile-quantile plot for determining the Normality of the distribution:
 Based on this graph, what is the best description of the data?

Accepted Answer

The data appears to follow a normal distribution, but it is not a perfect fit to the standard normal distribution line. There is a slight curve in the plot indicating that the tails of the distribution may be slightly heavier than those of a standard normal distribution.

Question 20

A researcher was interested in the survival times of brook trout after their exposure to increased levels of total dissolved solids. The following quantile-quantile plot was obtained:
 Based on this graph, what is the best description of the data?

Accepted Answer

The answer of A researcher was interested in the survival...

Quiz 10: The Normal Distributions

Serum albumin levels in adults are distributed approximately Normally with a mean of 35 g/l and a standard deviation of 5 g/l. What is the probability that a randomly selected adult has a serum albumin level greater than 32.5 g/l?

Serum albumin levels in adults are distributed approximately Normally with a mean of 35 g/l and a standard deviation of 5 g/l. What serum albumin level do the lower 10% of adults have?

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg. What is the approximate probability that a randomly selected man with liver cirrhosis has a total body protein of 9.75 kg or greater?

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg. What is the approximate probability that a randomly selected man with liver cirrhosis has a total body protein between 9.75 and 9.85 kg?

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg. What is the minimum total body protein that the upper 25% of adult men with cirrhosis have?

The distribution of total body protein in adult men with liver cirrhosis is approximately Normal with mean 9.8 kg and standard deviation 0.1 kg. We obtain the numerical value for P(X ≤ 9.65) . Which of the following statements about the probability, P(X < 9.65) , is true?

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg. What is the approximate probability that a randomly selected healthy man has a total body protein greater than 12 kg?

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg. What is the approximate probability that a randomly selected healthy man has a total body protein between 12.1 and 12.2 kg?

The distribution of total body protein in healthy adult men is approximately Normal with mean 12.3 kg and standard deviation 0.1 kg. What is the minimum total body protein for the upper 75% of healthy adult men?

The distribution of vitamin C amount in vitamin drops produced by a given factory is approximately Normal, with a mean of 60.0 mg and a standard deviation of 0.5 mg. What is the approximate proportion of vitamin drops containing more than 60.75 mg of vitamin C?

The distribution of vitamin C amount in vitamin drops produced by a given factory is approximately Normal, with a mean of 60.0 mg and a standard deviation of 0.5 mg. What is the approximate probability of drawing at random a vitamin drop with a vitamin content between 60.25 and 60.75 mg?

The amount of cholesterol in a person's body produced by the liver and other cells is proposed to be Normally distributed with mean 75% and standard deviation 0.5%. What is the probability that a person produces more than 76.7% of the cholesterol in his or her body?

The amount of cholesterol in a person's body produced by the liver and other cells is proposed to be Normally distributed with mean 75% and standard deviation 0.5%. What is the first quartile of the distribution of cholesterol production?

X has a Normal distribution with mean 50 and standard deviation 8. Which of the following statements is correct?

Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 micron. What position (in microns) are 20% of binding sites greater than?

Using fluorescent imaging techniques, researchers observed that the position of binding sites on HIV peptides is approximately Normally distributed with a mean of 2.45 microns and a standard deviation of 0.35 microns. What is the standardized score for a binding site position of 2.03 microns?

A researcher was interested in the lengths of brook trout and obtained the following quantile-quantile plot for determining the Normality of the distribution: Based on this graph, what is the best description of the data?

A researcher was interested in the survival times of brook trout after their exposure to increased levels of total dissolved solids. The following quantile-quantile plot was obtained: Based on this graph, what is the best description of the data?