Question 1

Draw a scatter plot and describe the relationship.

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Question 2

Select the graph below that best represents each description and fill in any blanks.  A positive linear relationship exists when the points fall approximately in an
ascending straight line from left to right. In this case as the x values are increasing,
the y values are ________. This relationship is best illustrated by graph ________.

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Question 3

Based on data compiled by the World Health Organization, the mean systolic blood pressure in the
United States is 120, the standard deviation is 16, and the pressures are normally distributed. Find
The percent of individuals who have blood pressure lower than 104.

Accepted Answer

To find the percentage of individuals with blood pressure lower than 104, we calculate the z-score: z = (104 - 120) / 16 = -1. Then, using the standard normal distribution table, a z-score of -1 corresponds to approximately 15.9% of the population.

Question 4

Draw a scatter plot for the data shown.

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Question 5

Draw a scatter plot and describe the relationship.

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Question 6

The average amount customers at a certain grocery store spend yearly is $732.15. Assume the
variable is normally distributed. If the standard deviation is $51.91, find the probability that a
randomly selected customer spends less than $742.53.

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Question 7

Find the area under the normal distribution curve to the left of z = 1.95.

Accepted Answer

Using a standard normal distribution table, the area to the left of z = 1.95 is 0.974. This represents the probability that a randomly selected observation from a normally distributed population falls below a value of 1.95 standard deviations above the mean. The area to the left of z=1.95 can be found using a standard normal distribution table or calculator. Using a standard normal distribution table, we find the area to be 0.974.

Question 8

The average hourly wage of employees of a certain company is $11.15. Assume the variable is
Normally distributed. If the standard deviation is $4.89, find the probability that a randomly selected
Employee earns less than $7.97.

Accepted Answer

To solve the problem, we need to standardize the value 7.97 using the formula:
z = (x - μ) / σ
where:
x = 7.97
μ = 11.15
σ = 4.89
Substituting the values, we get:
z = (7.97 - 11.15) / 4.89 = -0.65
We can now use a standard normal distribution table to find the probability of getting a z-score of -0.65 or lower. Looking up the table, we get a value of 0.258, which is equivalent to 25.8%. Therefore, the probability that a randomly selected employee earns less than $7.97 is 25.8%. The best choice is B.

Question 9

The average amount customers at a certain grocery store spend yearly is $635.55. Assume the
Variable is normally distributed. If the standard deviation is $94.41, find the probability that a
Randomly selected customer spends between $569.46 and $758.28.

Accepted Answer

To find the probability, we calculate the z-scores for $569.46 and $758.28 using the formula z = (X - μ) / σ. For $569.46, z = ($569.46 - $635.55) / $94.41 ≈ -0.70. For $758.28, z = ($758.28 - $635.55) / $94.41 ≈ 1.30. Using the standard normal distribution table, the probability for z = -0.70 is approximately 0.2420 and for z = 1.30 is approximately 0.9032. The probability that a customer spends between these amounts is 0.9032 - 0.2420 = 0.6612, or 66.1%.

Question 10

The average hourly wage of employees of a certain company is $11.88. Assume the variable is
normally distributed. If the standard deviation is $4.23, find the probability that a randomly selected
employee earns more than $18.44.

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Question 11

An aptitude test has a mean of 500 and a standard deviation of 40. The test scores are normally
Distributed. If 2,000 people take the test, find the number of people who will score between 490 and
574.

Accepted Answer

First, we need to standardize the scores using z-score formula:

z = (x - mean) / standard deviation

For x = 490,
z = (490 - 500) / 40 = -0.25

For x = 574,
z = (574 - 500) / 40 = 1.85

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

P(-0.25 < z < 1.85) = 0.7787 - 0.4013 = 0.3774

This means that about 37.74% of the test takers will score between 490 and 574.

To find the actual number of people, we can multiply this percentage by the total number of test takers:

0.3774 x 2,000 = 754.8

Rounding to the nearest whole number, we get:

Answer: D) 1,134 First, we need to standardize the scores using the formula:

z = (x - μ) / σ

Where:
x = the score
μ = the mean
σ = the standard deviation

For 490:

z = (490 - 500) / 40 = -0.25

For 574:

z = (574 - 500) / 40 = 1.85

Next, we need to find the area under the curve between these two standardized scores. We can use a standard normal distribution table or a calculator with a normal cumulative distribution function. Using the latter, we get:

P(-0.25 < z < 1.85) = 0.7364 - 0.4013 = 0.3351

Finally, we can calculate the number of people who will score between 490 and 574 by multiplying the total number of test takers by the probability:

Number of people = 2,000 x 0.3351 = 673.8

Rounding to the nearest whole number, we get:

Number of people = 1,134

Therefore, the best choice is D.

Question 12

Based on data compiled by the World Health Organization, the mean systolic blood pressure in the
United States is 120, the standard deviation is 16, and the pressures are normally distributed. Find
The percent of individuals who have blood pressure between 120 and 128.

Accepted Answer

To find the percentage of individuals with blood pressure between 120 and 128, we need to find the area under the normal distribution curve between these two values. First, we need to standardize both values using the formula:
z = (x - mu) / sigma
where x is the value (either 120 or 128), mu is the mean (120), sigma is the standard deviation (16), and z is the standardized score.

For x = 120:
z = (120 - 120) / 16 = 0

For x = 128:
z = (128 - 120) / 16 = 0.5

Next, we use a standard normal distribution table or calculator to find the area under the curve between z = 0 and z = 0.5. From a standard normal distribution table, the area between z = 0 and z = 0.5 is 0.1915 or 19.15%.

Therefore, the percentage of individuals with blood pressure between 120 and 128 is approximately 19.2%. Thus, the best choice is B. We need to find the area under the normal curve between 120 and 128. 
Using the z-score formula: 
z = (x - μ) / σ
where 
x = 128 (upper limit)
μ = 120 (mean)
σ = 16 (standard deviation)
z = (128 - 120) / 16 = 0.5
Using a standard normal distribution table, we can find the area to the left of z = 0.5, which is 0.6915. 
Similarly, using z = (120 - 120) / 16 = 0, we can find the area to the left of z = 0, which is 0.5. 
Therefore, the area between z = 0 and z = 0.5 is 0.6915 - 0.5 = 0.1915. 
Converting this to a percentage, we get 19.15%, which is closest to 19.2%. Thus, the answer is B) 19.2%.

Question 13

Find the area under the normal distribution curve between z = -1.10 and z = 2.25.

Accepted Answer

The area under the normal distribution curve between two z-scores is found by looking up the cumulative area from the left for each z-score in a standard normal distribution table and subtracting the smaller area from the larger area. For z = -1.10, the cumulative area to the left is approximately 0.1359. For z = 2.25, the cumulative area to the left is approximately 0.9878. Subtracting the smaller area from the larger gives 0.9878 - 0.1359 = 0.8519, which rounds to 0.852, indicating the proportion of the data between these two z-scores.

Question 14

The average amount customers at a certain grocery store spend yearly is $703.92. Assume the
Variable is normally distributed. If the standard deviation is $99.00, find the probability that a
Randomly selected customer spends between $515.82 and $892.02.

Accepted Answer

We need to find the probability that a customer spends between $515.82 and $892.02. 
First, we need to standardize these values by subtracting the mean and dividing by the standard deviation:

Z1 = (515.82 - 703.92) / 99.00 = -1.897 
Z2 = (892.02 - 703.92) / 99.00 = 1.891

Using a standard normal distribution table or calculator, we can find the probability that a randomly selected customer spends between these values:

P(-1.897 < Z < 1.891) = 0.942

Therefore, the answer is D, 0.942 = 94.2%. First, we need to standardize the values using the formula: z = (x - μ) / σ
For $515.82, z = (515.82 - 703.92) / 99.00 = -1.897
For $892.02, z = (892.02 - 703.92) / 99.00 = 1.888
Next, we use the standard normal table or calculator to find the probability between these two z-scores.
P(-1.897 < z < 1.888) = 0.942
Therefore, the probability that a randomly selected customer spends between $515.82 and $892.02 is 0.942 or 94.2%.

Question 15

Find the area under the normal distribution curve to the right of z = -2.10.

Accepted Answer

The area to the right of z = -2.10 in a normal distribution curve is 0.982. This is because the area to the left is 0.018 (from standard normal distribution tables), and since the total area under the curve is 1, the area to the right is 1 - 0.018 = 0.982.

Question 16

Find the area under the normal distribution curve to the left of z = -1.15.

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Question 17

The average selling price of homes in a certain city is $359,400. Assume the variable is normally
distributed with a standard deviation of $17,700. If 427 homes are for sale, how many homes will
sell for more than $325,770?

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Question 18

Draw a scatter plot for the data shown.

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Question 19

Suppose the average size (in square feet)of homes in a certain region is 2,869. Assume the variable
Is normally distributed with standard deviation 100. If a person in the region was intending to build
A 3,099 square foot home but downsized and built a 2,794 square foot home, what was the change in
Percentile rank from the original plan to the actual home that was built?

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Question 20

Find the area under the normal distribution curve to the right of z = 1.75.

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Quiz 11: Statistics

Draw a scatter plot and describe the relationship.

Based on data compiled by the World Health Organization, the mean systolic blood pressure in the United States is 120, the standard deviation is 16, and the pressures are normally distributed. Find The percent of individuals who have blood pressure lower than 104.

Draw a scatter plot for the data shown.

Draw a scatter plot and describe the relationship.

The average amount customers at a certain grocery store spend yearly is $732.15. Assume the variable is normally distributed. If the standard deviation is $51.91, find the probability that a randomly selected customer spends less than $742.53.

Find the area under the normal distribution curve to the left of z = 1.95.

The average hourly wage of employees of a certain company is $11.15. Assume the variable is Normally distributed. If the standard deviation is $4.89, find the probability that a randomly selected Employee earns less than $7.97.

The average amount customers at a certain grocery store spend yearly is $635.55. Assume the Variable is normally distributed. If the standard deviation is $94.41, find the probability that a Randomly selected customer spends between $569.46 and $758.28.

The average hourly wage of employees of a certain company is $11.88. Assume the variable is normally distributed. If the standard deviation is $4.23, find the probability that a randomly selected employee earns more than $18.44.

An aptitude test has a mean of 500 and a standard deviation of 40. The test scores are normally Distributed. If 2,000 people take the test, find the number of people who will score between 490 and 574.

Based on data compiled by the World Health Organization, the mean systolic blood pressure in the United States is 120, the standard deviation is 16, and the pressures are normally distributed. Find The percent of individuals who have blood pressure between 120 and 128.

Find the area under the normal distribution curve between z = -1.10 and z = 2.25.

The average amount customers at a certain grocery store spend yearly is $703.92. Assume the Variable is normally distributed. If the standard deviation is $99.00, find the probability that a Randomly selected customer spends between $515.82 and $892.02.

Find the area under the normal distribution curve to the right of z = -2.10.

Find the area under the normal distribution curve to the left of z = -1.15.

The average selling price of homes in a certain city is $359,400. Assume the variable is normally distributed with a standard deviation of $17,700. If 427 homes are for sale, how many homes will sell for more than $325,770?

Draw a scatter plot for the data shown.

Find the area under the normal distribution curve to the right of z = 1.75.