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In Exercises 37–40, 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

Blood Platelet Count of Males

Frequency

0-99

1

100-199

51

200-299

90

300-399

10

400-499

0

500-599

0

600-699

1

Short Answer

Expert verified

The standard deviation is computed as 68.4.

The value of 68.4 differs from the value of 59.5 by a significant amount. Therefore, these two values are not equal.

Step by step solution

01

Given information

The frequency distribution of the number of males corresponding to the blood platelet count is tabulated.

02

Formula

Thestandard deviationis a measure of dispersion most commonly used for a set of data. For a frequency distribution, the following formula is used:

s=nf×x2-f×x2nn-1

Here,

  • the frequencies are denoted by f;
  • the class midpoints are denoted by x;
  • the sum of frequencies is denoted by n.
03

Calculations

The calculations are tabulated below:

Blood Platelet Count

Frequency (f)

Midpoint

(x)

f×x

f×x2

0-99

1

49.5

49.5

2450.25

100-199

51

149.5

7624.5

1139863

200-299

90

249.5

22455

5602523

300-399

10

349.5

3495

1221503

400-499

0

449.5

0

0

500-599

0

549.5

0

0

600-699

1

649.5

649.5

421850.3


n=153

f×x=34273.5

f×x2=8388188

Substituting the above values in the formula:

s=nf×x2-f×x2nn-1=1538388188-34273.52153153-1=68.4

Therefore, the computed value of the standard deviation is68.4.

04

Compare the actual and computed value of standard deviation 

The actual value of standard deviation is 59.5.

The calculated value of the standard deviation equal to 68.4differs from the value obtained from the original list of observations, which is 59.5.

Thus, the two values cannot be considered equal.

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Most popular questions from this chapter

The Empirical Rule Based on Data Set 1 “Body Data” in Appendix B, blood platelet counts of women have a bell-shaped distribution with a mean of 255.1 and a standard deviation of 65.4. (All units are 1000 cells/L.) Using the empirical rule, what is the approximate percentage of women with platelet counts

a. within 2 standard deviations of the mean, or between 124.3 and 385.9?

b. between 189.7 and 320.5?

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?

In Exercises 21–28, use the same list of Sprint airport data speeds (Mbps) given for Exercises 17–20. Find the indicated percentile or quartile.

P40

In Exercises 21–28, use the same list of Sprint airport data speeds (Mbps) given for Exercises 17–20. Find the indicated percentile or quartile.

P60

Degrees of Freedom Five pulse rates randomly selected from Data Set 1 “Body Data” in Appendix B have a mean of 78.0 beats per minute. Four of the pulse rates are 82, 78, 56, and 84.

a. Find the missing value.

b. We need to create a list of n values that have a specific known mean. We are free to select any values we desire for some of the n values. How many of the n values can be freely assigned before the remaining values are determined? (The result is referred to as the number of degreesof freedom.)

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