26 64 36 42 70 84 46 51 49 (a) Calculate the variance and standard deviation for this data set. (b) If a very expensive cheese with a cost per slice of 150 cents was added to the data set, how would the values of the mean and standard deviation change? Step 1 (a) Calculate the variance and standard deviation for this data set. Recall the given data. 26 64 36 42 70 84 46 51 49 The first step in finding a sample's variance and standard deviation is to calculate the sample mean. Substitute the sample values into the formula and simplify to calculate the sample mean x. - 26 + 64 + 36 + 42 + 70 + 84 + 46 + 51 + 49 468 468 52 Step 2 We found the sample mean to be 52. The next step is to calculate the deviation from the mean for each sample value and square these deviations from the mean. Complete the following table. Squared Deviation from the Mean Cost (in cents) per ounce of sliced Swiss cheese Deviation from the Mean (x - x) (x - x)2 26 - 52 - -26 (26 - 52)2- 676 26 64 64 - 52 - 12 12 (64 - 52)- 144 144 36 -16 256 42 -10 100 70 18 324 84 32 1024 46 36 51 -1 49 The sample variance, s, is the sum of squared deviations from the mean divided by n - 1. n-1 The squared deviations were found previously as shown in the table below. Squared Deviation from the Mean Cost (in cents) per ounce of sliced Swiss cheese Deviation from the Mean (x - x) (x - x)2 26 -26 676 64 12 144 36 -16 256 42 -10 100 70 18 324 84 32 1024 46 -6 36 51 -1 49 -3 Since there are nine diffrent brands of Swiss cheese represented in the sample, n= 9v 9 Substitute n and the squared deviations into the formula and simplify to obtain the sample variance. n-1 676 + 144 + 256 + 100 + 324 + 1024 + 36 +1+9 9 -1 321.25 321.25 Step 4 The sample standard deviation, s, is the positive square root of the sample variance, s. We previously calculated the sample variance to be s - 321.25. Find the sample standard deviation. Round to four decimals. - V 321.25 321.25 17.92344833 17.9234 Step 5 (b) If a very expensive cheese with a cost per slice of 150 cents was added to the data set, how would the values of the mean and standard deviation change? Recall the original nine data values. 26 64 36 42 70 84 46 51 49 By including this additional cheese in our data set, our sample size changes. What is the new sample size? n -9+9 - 18 What is the new sample mean, x? 26 + 64 + 36 + 42 + 70 + 84 + 46 + 51 + 49 + 150 18 -34.33 We found the mean of the original data to be 52. The new mean with the addition of the very expensive cheese increases V the value of the mean.