Question 1

"The records of the National Weather Service indicate that, on the average, the city of San Jose receives 14.03 inches of rain each year, with a standard deviation of 5.19 inches." How many inches of rain must fall for a year to be in the top 10%?

Accepted Answer

To find the number of inches of rain that must fall for a year to be in the top 10%, we use the z-score formula associated with the top 10% of a normal distribution. The z-score for the top 10% is approximately 1.28. We then apply the formula: X = μ + zσ, where μ is the mean (14.03 inches), σ is the standard deviation (5.19 inches), and z is the z-score (1.28). Plugging in the values, we get X = 14.03 + (1.28)(5.19) ≈ 20.67 inches.

Question 2

"IQ scores based on the Wechsler Adult Intelligence Scale are assumed to be normally distributed with a mean of 100 and a standard deviation of 15." The label 'Genius' is given to people whose IQ scores are 2 standard deviations above the mean. What IQ score do people need to have to be labeled a 'Genius'?

Accepted Answer

Two standard deviations above the mean would be 2 x 15 = 30. Adding this to the mean of 100 gives an IQ score of 130 as the threshold for being labeled a 'Genius.'

Question 3

"A study examining people's water consumption estimated men spend an average of 10.4 minutes taking a shower, with a standard deviation of 4.8 minutes (Shimokura et al., 1998)." What percentage of men spend between 10 and 20 minutes taking a shower?

Accepted Answer

To find the percentage of men who spend between 10 and 20 minutes taking a shower, first we need to convert these values to z-scores using the formula: z = (x - μ) / σ where x is the time in minutes, μ is the mean time (10.4), and σ is the standard deviation (4.8).

For x = 10 minutes, z = (10 - 10.4) / 4.8 = -0.08. 
For x = 20 minutes, z = (20 - 10.4) / 4.8 = 1.96.

We can then use a standard normal distribution table or calculator to find the area (or percentage) under the curve between these two z-scores. This area represents the percentage of men who spend between 10 and 20 minutes taking a shower.

Using a calculator, we find that the area between z = -0.08 and z = 1.96 is approximately 0.5091, or 50.91%. Therefore, the correct answer is choice C.

Question 4

"According to the National Center for Health Statistics, in 2005 the average birthweight of a newborn baby was approximately normally distributed with a mean of 119 ounces (7 lb, 7 oz) and a standard deviation of 19 ounces." What percentage of babies weigh between 112 and 144 ounces (7 and 9 lb) at birth?

Accepted Answer

To find the percentage of babies weighing between 112 and 144 ounces, we convert these values to z-scores using the formula $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. For 112 ounces, $z = \frac{(112 - 119)}{19} = -0.37$, and for 144 ounces, $z = \frac{(144 - 119)}{19} = 1.32$. Looking these z-scores up in a standard normal distribution table, we find that the percentages are approximately 35.62% for -0.37 and 90.71% for 1.32. The difference, 90.71% - 35.62%, gives us approximately 55.09%, which represents the percentage of babies weighing between 112 and 144 ounces.

Question 5

"The records of the National Weather Service indicate that, on the average, the city of San Jose receives 14.03 inches of rain each year, with a standard deviation of 5.19 inches." In what percentage of years would you expect there to be less than 12 inches of rain?

Accepted Answer

To find the percentage of years with less than 12 inches of rain, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (12 inches), μ is the mean (14.03 inches), and σ is the standard deviation (5.19 inches). Plugging in the values: Z = (12 - 14.03) / 5.19 = -2.03 / 5.19 ≈ -0.39. Looking up -0.39 on a standard normal distribution table or using a calculator gives us an area to the left of approximately 34.83%, which represents the percentage of years with less than 12 inches of rain.

Question 6

"According to the National Center for Health Statistics, in 2005 the average birthweight of a newborn baby was approximately normally distributed with a mean of 119 ounces (7 lb, 7 oz) and a standard deviation of 19 ounces." A 'low birthweight' baby is one who weighs less than 88 ounces (5 lb, 8 oz) at birth. What percentage of babies are low birthweight?

Accepted Answer

To find the percentage of babies that are considered low birthweight, we use the Z-score formula: Z = (X - μ) / σ, where X is the value in question (88 ounces), μ is the mean (119 ounces), and σ is the standard deviation (19 ounces). Plugging in the values gives us Z = (88 - 119) / 19 = -31 / 19 ≈ -1.63. Looking up this Z-score in a standard normal distribution table or using a calculator, we find that the percentage of babies weighing less than 88 ounces is approximately 5.16%.

Question 7

"Imagine you are president of a toy company that builds video games. You have determined that the time needed to build these games is normally distributed with a mean of 20 minutes and a standard deviation of 5 minutes." What percentage of video games are built in LESS than 15 minutes?

Accepted Answer

To find the percentage of video games built in less than 15 minutes, we use the Z-score formula: Z = (X - μ) / σ, where X is the value (15 minutes), μ is the mean (20 minutes), and σ is the standard deviation (5 minutes). Plugging in the values, we get Z = (15 - 20) / 5 = -1. A Z-score of -1 corresponds to the left tail of the normal distribution, which is approximately 15.87% of the area under the curve.

Question 8

For a sample of raw scores, you calculate a mean of 15.25 and a standard deviation of 3.19. What is the z-score for a raw score of 11?

Accepted Answer

Z-score formula: z = (x - mean) / standard deviation

Plugging in the values:

z = (11 - 15.25) / 3.19
z = -1.33

Therefore, the answer is D.

Question 9

If a variable is normally distributed with a hypothesized mean (μ) of 26 and a standard deviation (σ) of 7, what is the z-score for a raw score (X) of 21?

Accepted Answer

To find the z-score, we use the formula:
z = (X-μ)/σ
Plugging in our values, we get:
z = (21-26)/7 = -0.71
Therefore, the z-score for a raw score of 21 is -0.71.

Question 10

IQ scores based on the Wechsler Adult Intelligence Scale are assumed to be normally distributed with a mean of 100 and a standard deviation of 15. What is the z-score for an IQ of 90?

Accepted Answer

To find the z-score, we use the formula:

z = (x - μ) / σ

Where x is the raw score (in this case, 90), μ is the mean (100), and σ is the standard deviation (15).

Plugging in the values, we get:

z = (90 - 100) / 15
z = -0.67

Therefore, the z-score for an IQ of 90 is -0.67, which corresponds to choice A.

Question 11

"According to the National Center for Health Statistics, in 2005 the average birthweight of a newborn baby was approximately normally distributed with a mean of 119 ounces (7 lb, 7 oz) and a standard deviation of 19 ounces." What percentage of babies weighed more than 128 ounces (8 lb) at birth?

Accepted Answer

The answer of "According to the National Center for Health...

Question 12

"IQ scores based on the Wechsler Adult Intelligence Scale are assumed to be normally distributed with a mean of 100 and a standard deviation of 15." What percentage of adults have IQ scores less than 90?

Accepted Answer

The answer of "IQ scores based on the Wechsler Adult...

Question 13

IQ scores based on the Wechsler Adult Intelligence Scale (WAIS) are assumed to be normally distributed with a mean of 100 and a standard deviation of 15. What is the z-score for an IQ of 125?

Accepted Answer

The formula for calculating z-score is (observed value - mean)/standard deviation. Plugging in the values, we get (125-100)/15 = 1.67. Therefore, the z-score for an IQ of 125 is 1.67, which corresponds to answer choice C.

Question 14

You can apply the normal curve table to a standardized distribution ______.

Accepted Answer

The normal curve table can be applied to a standardized distribution when the distribution of raw scores is normally distributed. The other options are not relevant to the use of a normal curve table.

Question 15

"Imagine you are president of a toy company that builds video games. You have determined that the time needed to build these games is normally distributed with a mean of 20 minutes and a standard deviation of 5 minutes." 5% of the video games (the bottom 5%) take less than ______ minutes to build.

Accepted Answer

To find the time for the bottom 5%, we need to look for the value that corresponds to a cumulative area of 0.05 in a normal distribution table. Using the Z-score formula, we get:
Z = (X - μ) / σ
-1.645 = (X - 20) / 5
X = 11.775
Rounding to the nearest whole number, we get 12 minutes, which is the closest option to our calculated value. Therefore, the answer is B.

Question 16

"IQ scores based on the Wechsler Adult Intelligence Scale are assumed to be normally distributed with a mean of 100 and a standard deviation of 15." What percentage of adults have IQ scores between 95 and 110?

Accepted Answer

The percentage of adults with IQ scores between 95 and 110 can be calculated using the normal distribution formula:P(95 < X < 110) = P(Z < (110 - 100) / 15) - P(Z < (95 - 100) / 15)= P(Z < 0.67) - P(Z < -0.33)= 0.7486 - 0.3707= 0.3779Therefore, the percentage of adults with IQ scores between 95 and 110 is 37.79%.

Question 17

"A recent study of Greek children estimated that the average weight of a 10-year old boy was 81.57 pounds with a standard deviation of 17.41 pounds (Papadimitriou et al., 2006)." What percentage of boys weigh more than 105 pounds?

Accepted Answer

To solve this problem, we need to use the standard normal distribution. We can convert the data to z-scores using the formula:

z = (x - μ) / σ

where x is the weight we are interested in (105 pounds), μ is the mean weight (81.57 pounds), and σ is the standard deviation (17.41 pounds).

z = (105 - 81.57) / 17.41 = 1.34

We can use a standard normal distribution table or calculator to find the probability of a z-score being greater than 1.34. This probability is approximately 0.0885, or 8.85%. Therefore, the answer is B) 8.85%.

Question 18

An instructor transforms her students' scores on a midterm into standardized (z) scores. Which of the following is by definition true about the distribution of standardized scores?

Accepted Answer

By definition, z scores transform raw scores into a standard normal distribution with a mean of 0 and a standard deviation of 1. Therefore, the mean of the standardized scores will always be 0. While the distribution may often be normally distributed, this is not always guaranteed, so choice A is incorrect. The distribution does stretch to infinity in both directions (i.e., it is unbounded), so choice C is partially correct. However, the standard deviation of the standardized scores is always 1, not equal to σ (the standard deviation of the original scores), so choice D is incorrect.

Question 19

"A study examining people's water consumption estimated men spend an average of 10.4 minutes taking a shower, with a standard deviation of 4.8 minutes (Shimokura et al., 1998)." What percentage of men spend more than 15 minutes taking a shower?

Accepted Answer

To find the percentage of men who spend more than 15 minutes taking a shower, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (15 minutes), μ is the mean (10.4 minutes), and σ is the standard deviation (4.8 minutes). Plugging in the values, we get Z = (15 - 10.4) / 4.8 = 0.9583. Looking up this Z-score in a standard normal distribution table or using a calculator, we find that the area to the right of this Z-score (which represents the percentage of men who spend more than 15 minutes) is approximately 16.85%.

Question 20

"A recent study of Greek children estimated that the average weight of a 10-year old boy was 81.57 pounds with a standard deviation of 17.41 pounds (Papadimitriou et al., 2006)." What percentage of boys weigh between 70 and 90 pounds?

Accepted Answer

First, we need to find the z-scores for the given weights:

z1 = (70-81.57)/17.41 = -0.66 
z2 = (90-81.57)/17.41 = 0.48

Using a standard normal distribution table or calculator, we can find the area under the curve between these z-scores:

area = P(-0.66 < z < 0.48)

area = P(z < 0.48) - P(z < -0.66)

area = 0.6844 - 0.2614

area = 0.423

Converting the decimal to a percentage, we get:

percentage = 42.98%

Therefore, the correct answer is (C) 42.98%.

Quiz 5: Normal Distributions