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In Exercises 5鈥20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as 鈥渕inutes鈥) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Football Player Weights Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII (the same players from the preceding exercise). Are the measures of variation likely to be typical of all NFL players?

189 254 235 225 190 305 195 202 190 252 305

Short Answer

Expert verified

The different measures of variation for the given sample are as follows:

  • The range is equal to 116.0 pounds.
  • The variance is equal to 1923.7 pounds squared.
  • The standard deviation is equal to 43.9 pounds.

The measure of variation cannot be considered typical of all NFL players as the sample is selected from the same team.

Step by step solution

01

Given information

Weights (in pounds) of 11 players (represented as n) are provided.

02

Measures of variation

The following are the measures of dispersion that are generally computed for a sample of data:

The difference between the greatest and the smallest values for a given set of data is called therange.

The ratio of the squared difference of a set of data from its mean to the value of the sample size minus one is called thesample variance.It can be mathematically written as

s2=1=1nxi-x2n-1

The under root value of the sample variance is same as the value of the sample standard deviation with the original units.

03

Computation of the range, variance, and standard deviation of the sample data 

The range of weights is obtained as

Range=GreatestValue-SmallestValue=305-189=116.0pounds

Therefore, the range of weights is equal to 116.0 pounds.

The mean weight of the players is calculated as

x=1=1nxin=189+254+...+30511=254211231.1

.

Thus, the mean weight of the players is 231.1 pounds.

The variance s2of the sample of weights is

s2=1=1nxi-x2n-1=189-231.12+254-231.12+...+305-231.1211-1=19236.909101923.7

Therefore, the variance of weights is equal to 1923.7 pounds squared.

The standard deviation of the sample of weights is equal to

s=s2=1923.743.9

Therefore, the standard deviation of weights is equal to 43.9 pounds.

04

Interpretation

Here, the players are chosen randomly from the same team (Seattle Seahawks). So, there is a possibility that the players of this team may possess the same features in common.

So, the calculated values cannot be used to represent the entire population of players.

Therefore, the measures of variation are not likely to be considered as the typical of all NFL players.

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Most popular questions from this chapter

In Exercises 5鈥20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as 鈥渕inutes鈥) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

TV Prices Listed below are selling prices in dollars of TVs that are 60 inches or larger and rated as a 鈥渂est buy鈥 by Consumer Reports magazine. Are the measures of variation likely to be typical for all TVs that are 60 inches or larger?

1800 1500 1200 1500 1400 1600 1500 950 1600 1150 1500 1750

In Exercises 33鈥36, use the range rule of thumb to identify the limits separating values that are significantly low or significantly high

Foot Lengths Based on Data Set 2 鈥淔oot and Height鈥 in Appendix B, adult males have foot lengths with a mean of 27.32 cm and a standard deviation of 1.29 cm. Is the adult male foot length of 30 cm significantly low or significantly high? Explain.

In Exercises 37鈥40, refer to the frequency distribution in the given exercise and find the standard deviation by using the formula below, where x represents the class midpoint, f represents the class frequency, and n represents the total number of sample values. Also, compare the computed standard deviations to these standard deviations obtained by using Formula 3-4 with the original list of data values: (Exercise 37) 11.5 years; (Exercise 38) 8.9 years; (Exercise 39) 59.5; (Exercise 40) 65.4.

Standard deviation for frequency distribution

s=nf.x2-f.x2nn-1

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Frequency

20-29

29

30-39

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3

60-69

5

70-79

1

80-89

1

In Exercises 9 鈥12, consider a value to be significantly low if its z score is less than or equal to -2 or consider the value to be significantly high if its z score is greater than or equal to 2.

Quarters Data Set 29 鈥淐oin Weights鈥 lists weights (grams) of quarters manufactured after 1964. Those weights have a mean of 5.63930 g and a standard deviation of 0.06194 g. Identify the weights that are significantly low or significantly high.

Resistant Measures Here are four of the Verizon data speeds (Mbps) from Figure 3-1: 13.5, 10.2, 21.1, 15.1. Find the mean and median of these four values. Then find the mean and median after including a fifth value of 142, which is an outlier. (One of the Verizon data speeds is 14.2 Mbps, but 142 is used here as an error resulting from an entry with a missing decimal point.) Compare the two sets of results. How much was the mean affected by the inclusion of the outlier? How much is the median affected by the inclusion of the outlier?

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