Explanation Definition: 1. While computing the mean of a frequency distribution, write all products x ⋅ f of the values and their frequencies in a new column of the table. Represent the sum of these products by ∑ ( x ⋅ f ) . Denote the sum of the frequencies by ∑ f . The mean is then ∑ ( x ⋅ f ) ∑ f . 2. If x is a data value in a set whose mean is x ¯ , then x − x ¯ is called x’s deviation from the mean. 3. We denote the standard deviation of a sample of n data values by s = ∑ ( x − x ¯ ) 2 n − 1 . 4. We calculate the standard deviation, s, of a sample that is given as a frequency distribution as s = ∑ ( x − x ¯ ) 2 ⋅ f n − 1 . Here f is the frequency of data value x , and n = ∑ f is the number of data values in the distribution. Calculation: From the given table understand each data value. x f 47 2 48 1 49 2 50 1 51 2 First find the mean of the given data set. To compute the mean, let us find the sum of given data values. Begin by finding the sum of the given data. Sum of 47, 48, 50, 49, 51, 47, 49, 51 is 392. There are 8 values in the given data. Therefore, the mean is x ¯ = 392 8 = 49. Now find the standard deviation. Observe that there are some repeated values in the given data. Therefore, we will use a frequency table to compute the standard deviation. The frequency of 47 is 2. That is, 47 come twice in the data set. Similarly, the frequency of 48 is 1. The frequency of 49 is 2. The frequency of 50 is 1. The frequency of 51 is 2. Now find the product x ⋅ f for each data value. Notice that 47 ⋅ 2 = 94 , 48 ⋅ 1 = 48 , 49 ⋅ 2 = 98 , 50 ⋅ 1 = 50 , and 52 ⋅ 1 = 102. The deviation is the difference between each product value and the mean. Therefore, the deviation of 47 from the mean 49 is 47 − 49 = − 2. The deviation of 48 from the mean 49 is 48 − 49 = − 1. The deviation of 49 from the mean 49 is 49 − 49 = 0. The deviation of 50 from the mean 49 is 50 − 49 = 1. The deviation of 51 from the mean 49 is 51 − 49 = 2. Square each of the deviation value

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ISBN: 9780134434681

Author: Tom Pirnot

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Chapter 14.3, Problem 25E

To determine

To find:

The standard deviation of H’s income from the mean using the following table:

Family A B C D E F G H

Annual income ($ thousands) 47 48 50 49 51 47 49 51

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Chapter 14.3, Problem 24E

Chapter 14.3, Problem 26E

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= 12:02 WeBWorK / 2024 Fall Rafeek MTH23 D02 / 9.2 Testing the Mean mu / 3 38 WEBWORK Previous Problem Problem List Next Problem 9.2 Testing the Mean mu: Problem 3 (1 point) Test the claim that the population of sophomore college students has a mean grade point average greater than 2.2. Sample statistics include n = 71, x = 2.44, and s = 0.9. Use a significance level of a = 0.01. The test statistic is The P-Value is between : The final conclusion is < P-value < A. There is sufficient evidence to support the claim that the mean grade point average is greater than 2.2. ○ B. There is not sufficient evidence to support the claim that the mean grade point average is greater than 2.2. Note: You can earn partial credit on this problem. Note: You are in the Reduced Scoring Period. All work counts for 50% of the original. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. . Oli wwm01.bcc.cuny.edu

>tt 1:32 > trend.1m 1m (sales > summary(trend.1m) - tt) #3###23 (i) #### Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2107.220 57.997 36.332e-16 *** tt -43.500 3.067 -14.18 7.72e-15 *** > trend = ts (fitted (trend.1m), start-start (sales), freq-frequency (sales)) sales trend ###23%23 (ii) #### as.numeric((1:32 %% 4) > X > q1 > q2 > q3 > 94 = = = = - as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) == 1) 2) == == 3) == 0) > season.lm = 1m (resid (trend.1m) 0+q1 + q2 + q3 + q4) #3##23%23 (iii) #### > summary(season.1m) Coefficients: Estimate Std. Error t value Pr(>|t|) q1 -38.41 43.27 -0.888 0.38232 92 18.80 43.27 0.435 0.66719 q3 -134.78 43.27 -3.115 0.00422 ** 94 154.38 43.27 3.568 0.00132 ** > season = ts (fitted (season.lm), start=start (sales), freq=frequency (sales)) > Y X season %23%23%23%23 (iv) #### >ar (Y, aic=FALSE, order.max=1) #23%23%23%23 (v) #### Coefficients: 1 0.5704 Order selected 1 sigma 2 estimated as 9431 > ar(Y, aic=FALSE,…

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Chapter 14 Solutions

Mathematics All Around (6th Edition)

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The standard deviation of H’s income from the mean using the following table: Family A B C D E F G H Annual income ($ thousands) 47 48 50 49 51 47 49 51

To find: The standard deviation of H’s income from the mean using the following table: Family A B C D E F G H Annual income ($ thousands) 47 48 50 49 51 47 49 51

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Chapter 14 Solutions

To find:

The standard deviation of H’s income from the mean using the following table:

Family A B C D E F G H

Annual income ($ thousands) 47 48 50 49 51 47 49 51