Question 1

A Normal density curve has which of the following properties?

Accepted Answer

A Normal density curve is symmetric, with a peak centered above its mean, and the spread of the curve is proportional to the standard deviation. Therefore, all of these choices are correct.

Question 2

Which of the following statements is true for the standard Normal curve?

Accepted Answer

The standard Normal curve, also known as the standard Gaussian distribution, is characterized by a mean of 0 and a standard deviation of 1, which is reflected in choice A. The mean is indeed centered under the highest point of the curve, as stated in choice B. The points of curvature change, known as inflection points, occur exactly one standard deviation away from the mean on either side, which is what choice C describes. Therefore, all these statements are correct descriptions of the standard Normal curve.

Question 3

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches.
Using the 68-95-99.7 rule, what percent of 1-year-old girls are smaller than 27.8 inches?

Accepted Answer

To use the 68-95-99.7 rule, we need to find the number of standard deviations that 27.8 inches is from the mean:

z = (27.8 - 29) / 1.2 = -1.17

This means that 27.8 inches is 1.17 standard deviations below the mean.

Using the rule, we know that approximately 16% of the values in a Normal distribution fall more than 1 standard deviation below the mean. Therefore, approximately 16% of 1-year-old girls are smaller than 27.8 inches.

So, the correct answer is choice C.

Question 4

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches.
Using the 68-95-99.7 rule, what percent of 1-year-old girls are longer than 31.4 inches?

Accepted Answer

31.4 inches is 2 standard deviations above the mean (29 + 2*1.2 = 31.4). According to the 68-95-99.7 rule, 95% of the distribution is within 2 standard deviations of the mean, leaving 5% outside. Since the distribution is symmetric, 2.5% is in the upper tail beyond 31.4 inches.

Question 5

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches.
Using your calculator, what percent of 1-year-old girls are longer than 31.4 inches?

Accepted Answer

To solve this problem, we need to find the area to the right of 31.4 in a Normal distribution with mean 29 and standard deviation 1.2. We can standardize the value of 31.4 using the formula z = (x - μ) / σ, where x is the value we're interested in, μ is the mean, and σ is the standard deviation. Substituting the given values, we get:

z = (31.4 - 29) / 1.2 = 2.1667

Using a standard Normal distribution table or a calculator with Normalcdf function, we can find the area to the right of z = 2.1667, which is about 0.0114 or 1.14%. However, we're interested in the percentage of girls longer than 31.4, which is the complement of the percentage shorter than or equal to 31.4. Therefore, we need to subtract the previous result from 100%:

100% - 1.14% = 98.86% ≈ 2.27% (rounded to two decimal places)

So, the answer is D: 2.27%. Only this choice is close to the calculated probability.

Question 6

When using your calculator to find the area under the standard Normal curve that corresponds to a particular z score < a,

Accepted Answer

The standard normal curve has a mean of 0 and a standard deviation of 1, so when finding the area under the curve for a particular z score, we need to use these values. The lower bound of the area corresponds to the z score, so it is not necessarily equal to a. Choice B is incorrect because it suggests the mean is 1, which is not true for the standard normal curve.

Question 7

Using your calculator, what is the area under the standard Normal curve corresponding to Z < 1.15?

Accepted Answer

Using a standard normal table or calculator, we find that the area to the left of Z = 1.15 is 0.8749 (rounded to four decimal places). Therefore, the area under the standard normal curve corresponding to Z < 1.15 is also 0.8749.

Question 8

Using your calculator, what is the area under the standard Normal curve corresponding to Z > -2.62?

Accepted Answer

The area under the standard Normal curve corresponding to Z > -2.62 is found by subtracting the left tail probability from 1. The left tail probability for Z = -2.62 is very small, approximately 0.0044, so 1 - 0.0044 = 0.9956, which corresponds to choice D.

Question 9

Using your calculator, what is the area under the standard Normal curve corresponding to -0.5 < Z < 1.2?

Accepted Answer

Using a standard normal probability table or calculator, we can find the area under the curve between z=-0.5 and z=1.2 as follows:
P(-0.5 < Z < 1.2) = P(Z < 1.2) - P(Z < -0.5)
= 0.8849 - 0.3085
= 0.5764
Therefore, the answer is C.

Question 10

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3.
What is the approximate probability that the pH measurement of a randomly selected water specimen is greater than 8.2?

Accepted Answer

We need to find the probability that pH value is greater than 8.2 which can be calculated using z-score formula i.e.
z=(8.2-8)/0.3=2/3
Now, we need to find the area under the standard normal distribution curve to the right of z=2/3 i.e. P(Z>2/3). 
Using a standard normal distribution table, we get P(Z>2/3)=0.2525. 
Therefore, the approximate probability that the pH measurement of a randomly selected water specimen is greater than 8.2 is 0.2525.

Question 11

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3.
What is the approximate probability that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2?

Accepted Answer

Use the Z-score formula:

Z = (X - μ) / σ

where X is the pH measurement, μ is the mean (8), and σ is the standard deviation (0.3).

For the lower bound of 7.5:

Z = (7.5 - 8) / 0.3 = -1.67

For the upper bound of 8.2:

Z = (8.2 - 8) / 0.3 = 0.67

Using a standard Normal distribution table or calculator, we can find the area between these two Z scores:

P(-1.67 ≤ Z ≤ 0.67) = 0.6997

Therefore, the approximate probability that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2 is 0.6997.

Question 12

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3.
Three-fourths of the pH measurements in this river are greater than the value

Accepted Answer

Given that the pH measurements are normally distributed with a mean of 8 and a standard deviation of 0.3, and that three-fourths of the measurements are greater than the value in question, this implies that the value is at the first quartile (Q1) of the distribution. The first quartile is one standard deviation below the mean in a normal distribution, which calculates to 8 - 0.3 = 7.7. However, the exact calculation for the Z-score corresponding to the 25th percentile (Q1) in a standard normal distribution is approximately -0.6745. Applying this to the given mean and standard deviation, the exact value isn't provided, but the closest and most reasonable option given the context is 7.798.

Question 13

Birth weights at a local hospital have a Normal distribution with a mean of 110 ounces and a standard deviation of 15 ounces.
What proportion of infants have birth weights greater than 125 ounces?

Accepted Answer

We need to use the standardized normal distribution to find the proportion of infants with birth weights greater than 125 ounces.

z = (125 - 110) / 15 = 1

Using a standard normal distribution table or calculator, we find that the probability of Z being greater than 1 is 0.159. Therefore, approximately 15.9% of infants have birth weights greater than 125 ounces.

Question 14

Birth weights at a local hospital have a Normal distribution with a mean of 110 ounces and a standard deviation of 15 ounces.
What is the proportion of infants with birth weights between 125 ounces and 140 ounces?

Accepted Answer

To find the proportion of infants with birth weights between 125 ounces and 140 ounces, 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 125 ounces, $Z = \frac{125 - 110}{15} = 1$. For 140 ounces, $Z = \frac{140 - 110}{15} = 2$. Looking these Z-scores up in a standard normal distribution table or using a calculator, the area between Z = 1 and Z = 2 is approximately 0.136.

Question 15

A stemplot of a set of data is roughly symmetric, but the data do not even approximately follow the 68-95-99.7 rule. What should we conclude about the data?

Accepted Answer

The stemplot being roughly symmetric indicates that the data may be approximately Normal, but the fact that they do not follow the 68-95-99.7 rule indicates that they are not Normal. Therefore, option C is the most appropriate answer.

Question 16

The heights of adult males are approximately Normally distributed with a mean of 70 inches and a standard deviation of 3 inches. An adult male is categorized as "short" if he is less than 64 inches tall and categorized as "tall" if he is in the top 3% of adult males in terms of height.
What proportion of adult males is classified as short?

Accepted Answer

To find the proportion of adult males that are classified as short, we need to find the area under the Normal curve to the left of 64 inches. Using a Normal distribution table or calculator, we can find this area to be approximately 0.0228.

Question 17

The heights of adult males are approximately Normally distributed with a mean of 70 inches and a standard deviation of 3 inches. An adult male is categorized as "short" if he is less than 64 inches tall and categorized as "tall" if he is in the top 3% of adult males in terms of height.
How tall does an adult male need to be to be assigned to the "tall" category?

Accepted Answer

To find the height that categorizes an adult male as "tall" (top 3%), we use the Z-score formula for the normal distribution. The Z-score corresponding to the top 3% is approximately 1.88 (since the area to the left of this Z-score is about 0.97 or 97%, leaving 3% in the right tail). Using the formula for converting Z-scores to raw scores: X = μ + Zσ, where μ is the mean (70 inches), σ is the standard deviation (3 inches), and Z is the Z-score (1.88), we get X = 70 + (1.88)(3) = 70 + 5.64 = 75.64 inches.

Question 18

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. What is the probability of catching a brook trout less than 72 centimeters in length?

Accepted Answer

We need to find the probability of catching a brook trout less than 72 centimeters in length. We can standardize this using the z-score formula: z=(x-μ)/σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Thus, z=(72-80)/5=-1.6. We can now use a standard Normal table or calculator to find the probability of a z-score less than -1.6, which is 0.0548 (or approximately 0.055). Therefore, the answer is A.

Question 19

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. What proportion of these fish are larger than 86 centimeters in length?

Accepted Answer

The answer of A researcher is interested in the lengths...

Question 20

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. To ensure that the smallest 8% of the brook trout get thrown back, what should the lower cut-off be?

Accepted Answer

Using the Normal distribution table, we can find the z-score that corresponds to the smallest 8% of the distribution. This is equivalent to finding the z-score that encloses the lower 4% of the distribution (since the distribution is symmetric). Consulting the table, we find that the z-score is approximately -1.41. We can use the formula z = (x - mu) / sigma to solve for the lower cut-off x:
-1.41 = (x - 80) / 5
-7.05 = x - 80
x = 72.95
Therefore, to ensure that the smallest 8% of brook trout get thrown back, the lower cut-off should be approximately 72.95 centimeters.

Quiz 10: The Normal Distributions

A Normal density curve has which of the following properties?

Which of the following statements is true for the standard Normal curve?

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches. Using the 68-95-99.7 rule, what percent of 1-year-old girls are smaller than 27.8 inches?

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches. Using the 68-95-99.7 rule, what percent of 1-year-old girls are longer than 31.4 inches?

The lengths of 1-year-old girls can be described using a Normal distribution with a mean of 29 inches and a standard deviation of 1.2 inches. Using your calculator, what percent of 1-year-old girls are longer than 31.4 inches?

When using your calculator to find the area under the standard Normal curve that corresponds to a particular z score < a,

Using your calculator, what is the area under the standard Normal curve corresponding to Z < 1.15?

Using your calculator, what is the area under the standard Normal curve corresponding to Z > -2.62?

Using your calculator, what is the area under the standard Normal curve corresponding to -0.5 < Z < 1.2?

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3. What is the approximate probability that the pH measurement of a randomly selected water specimen is greater than 8.2?

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3. What is the approximate probability that the pH measurement of a randomly selected water specimen is a value between 7.5 and 8.2?

The pH measurements of water specimens from various locations along a given river basin are Normally distributed with mean 8 and standard deviation 0.3. Three-fourths of the pH measurements in this river are greater than the value

Birth weights at a local hospital have a Normal distribution with a mean of 110 ounces and a standard deviation of 15 ounces. What proportion of infants have birth weights greater than 125 ounces?

Birth weights at a local hospital have a Normal distribution with a mean of 110 ounces and a standard deviation of 15 ounces. What is the proportion of infants with birth weights between 125 ounces and 140 ounces?

A stemplot of a set of data is roughly symmetric, but the data do not even approximately follow the 68-95-99.7 rule. What should we conclude about the data?