Statistics Flashcards by Chrissy Kay

31% of American adults surveyed stated that giving up their phone for a day would be more difficult than giving up their significant other.
In 2012, the United States population over the age of 18 was approximately 240,210,000. According to recent research from the Pew Foundation, 91% of all American adults have cell phones of some type.
Use the statistics given to determine the approximate number of American adults who would find giving up their phone more difficult than giving up their partner.

Approximately _____ American adults would find giving up their phone more difficult than giving up their partner.
(Round to the nearest whole number as needed.)

67,763,241

How well did you know this?

Not at all

Perfectly

Fifty million of 94 million households watched the last episode of a popular television show. What percentage of households were not watching the show?

Approximately ____% of households were not watching the last episode of the show.

46.8%

How well did you know this?

Not at all

Perfectly

A sample obtained in such a way that every element in the population has an equal chance of being selected is called a/an _______ sample.

random sample

How well did you know this?

Not at all

Perfectly

If data values are listed in one column and the adjacent column indicates the number of times each value occurs, the data presentation is called a _______.

Frequency Distribution

How well did you know this?

Not at all

Perfectly

The government of a large city needs to determine whether the city’s residents will support the construction of a new jail. The government decides to conduct a survey of a sample of the city’s residents. Which one of the following procedures would be most appropriate for obtaining a sample of the city’s residents?

a) Survey a random sample of the employees and inmates at the old jail.
b) Survey every fifth person who walks into City Hall on a given day.
c) Survey a random sample of persons within each geographic region of the city.
d) Survey the first 200 people in an alphabetical online listing of the city’s residents.

How well did you know this?

Not at all

Perfectly

find the mean
7, 4, 3, 2, 8, 5, 1, 3

4.125

How well did you know this?

Not at all

Perfectly

find the mean
91, 95, 99, 97, 93, 95

How well did you know this?

Not at all

Perfectly

find the mean
100, 40, 70, 40, 60

How well did you know this?

Not at all

Perfectly

find the mean
1.6, 3.8, 5.0, 2.7, 4.2, 4.2, 3.2, 4.7, 3.6, 2.5, 2.5

~3.45

How well did you know this?

Not at all

Perfectly

find the mean for the data items in the given frequency distribution

~4.71

How well did you know this?

Not at all

Perfectly

find the mean for the data items in the given frequency distribution

~6.26

How well did you know this?

Not at all

Perfectly

find the median
7, 4, 3, 2, 8, 5, 1, 3

3.5

How well did you know this?

Not at all

Perfectly

find the median
91, 95, 99, 97, 93, 95

How well did you know this?

Not at all

Perfectly

find the median
100, 40, 70, 40, 60

How well did you know this?

Not at all

Perfectly

find the median
1.6, 3.8, 5.0, 2.7, 4.2, 4.2, 3.2, 4.7, 3.6, 2.5, 2.5

3.6

How well did you know this?

Not at all

Perfectly

Find the median for the data items in the frequency distribution

How well did you know this?

Not at all

Perfectly

Find the median for the data items in the frequency distribution

How well did you know this?

Not at all

Perfectly

find the mode
7, 4, 3, 2, 8, 5, 1, 3

How well did you know this?

Not at all

Perfectly

find the mode
91, 95, 99, 97, 93, 95

How well did you know this?

Not at all

Perfectly

find the mode
100, 40, 70, 40, 60

How well did you know this?

Not at all

Perfectly

find the mode
1.6, 3.8, 5.0, 2.7, 4.2, 4.2, 3.2, 4.7, 3.6, 2.5, 2.5

2.5, 4.2 (bimodal)

Find the mode for the data items in the frequency distribution in

Find the mode for the data items in the frequency distribution in

find the midrange
7, 4, 3, 2, 8, 5, 1, 3

4.5

find the midrange 91, 95, 99, 97, 93, 95

find the midrange 100, 40, 70, 40, 60

find the midrange 1.6, 3.8, 5.0, 2.7, 4.2, 4.2, 3.2, 4.7, 3.6, 2.5, 2.5

3.3

Find the midrange for the data items in the frequency distribution in

4.5

Find the midrange for the data items in the frequency distribution in

5.5

find the range 1, 2, 3, 4, 5

find the range 7, 9, 9, 15

find the range 3, 3, 4, 4, 5, 5

Find the standard deviation for the group of data items. Round answers to two decimal places. 1, 2, 3, 4, 5

~1.58

Find the standard deviation for the group of data items. Round answers to two decimal places. 7, 9, 9, 15

~3.46

Find the standard deviation for the group of data items. Round answers to two decimal places. 3, 3, 4, 4, 5, 5

~0.89

Find the standard deviation for the group of data items. Round answers to two decimal places. 1, 1, 1, 4, 7, 7, 7

Find the standard deviation for the group of data items. Round answers to two decimal places. 9, 5, 9, 5, 9, 5, 9, 5

~2.14

12.4 (1) The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. Find the score that is 1 standard deviation above the mean.

120

3 The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. Find the score that is 3 standard deviations above the mean

160

5 The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. Find the score that is 2.5 standard deviations above the mean

150

7 The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. Find the score that is 2 standard deviations below the mean.

9 The scores on a test are normally distributed with a mean of 100 and a standard deviation of 20. Find the score that is one-half a standard deviation below the mean.

Prices of a New Car (Q 11 - 21) Not everyone pays the same price for the same model of a car. The figure illustrates a normal distribution for the prices paid for a particular model of a new car. The mean is $27,000 and the standard deviation is $500. A) between $26,500 and $27,500 B) between $27,000 and $27,500 C) between $26,000 and $27,000 D) between $25,500 and $27,000 E) more than $27,500 F) less than $26,000

A) 68 B) 34 C) 47.5 D) 49.85 E) 16 F) 2.5

Intelligence quotients (IQs) on the Stanford-Binet intelligence test are normally distributed with a mean of 100 and a standard deviation of 16. Use the 68–95–99.7 Rule to find the percentage of people with IQs A) between 68 and 132. B) between 68 and 100 C) above 116 D) below 68 E) above 148

A) 95 B) 47.5 C) 16 D) 2.5 E) 0.15

A set of data items is normally distributed with a mean of 60 and a standard deviation of 8. Convert each data item to a z-score. A) 68 B) 84 C) 64 D) 74 E) 60 F) 52

A) 1 B) 3 C) 0.5 D) 1.75 E) 0 F) -1

12.5 (1-29 odd) Find the percentage of data items in a normal distribution that lie: a. below and b. above the given z-score. 1) z = 0.6 3) z = 1.2 5) z = -0.7 7) z = -1.2

1) 72.57% 27.43% 3) 88.49% 11.51% 5) 24.2% 75.8% 7) 11.51% 88.49%

Find the percentage of data items in a normal distribution that lie between 9) z = 0.2 & z = 1.4 11) z = 1 & z = 2.1 13) z = -1.5 & z = 1.5 15) z = -2 & z = -0.5

9) 33.99% 11) 15.74% 13) 86.64% 15) 28.57%

Blood Pressure Systolic blood pressure readings are normally distributed with a mean of 121 and a standard deviation of 15. (A reading above 140 is considered to be high blood pressure.) In Exercises 17–26, begin by converting any given blood pressure reading or readings into z-scores. Then use Table 12.17 to find the percentage of people with blood pressure readings a) below 142. b) above 130. c) above 103 d) between 142 and 154 e) between 112 and 130

a) 91.92% b) 27.43% c) 88.49% d) 6.69% e) 45.14%

I’m using a table showing z-scores and percentiles that has positive percentiles corresponding to positive z-scores and negative percentiles corresponding to negative z-scores. Does this make sense? Why?

does not make sense

My table showing z-scores and percentiles does not display the percentage of data items greater than a given value of z. Does this make sense? Why?

makes sense