Question

exercises 4.17 - 4.33
4.17 ● the following data are costs (in cents) per ounce for nine different brands of sliced swiss cheese (consumerreports.org):
29 62 37 41 70 82 47 52 49
a. calculate the variance and standard deviation for this data set. (hint: see example 4.8.)
b. if a very expensive cheese with a cost per slice of $1.50 (150 cents) was added to the data set, how would the values of the mean and standard deviation change?

Explanation:

Step1: Calculate the mean

The data set is \(29,62,37,70,82,47,52,49\).
The mean \(\bar{x}=\frac{29 + 62+37+70+82+47+52+49}{8}=\frac{428}{8} = 53.5\)

Step2: Calculate the squared - differences

\((29 - 53.5)^2=(-24.5)^2 = 600.25\), \((62 - 53.5)^2=(8.5)^2 = 72.25\), \((37 - 53.5)^2=(-16.5)^2 = 272.25\), \((70 - 53.5)^2=(16.5)^2 = 272.25\), \((82 - 53.5)^2=(28.5)^2 = 812.25\), \((47 - 53.5)^2=(-6.5)^2 = 42.25\), \((52 - 53.5)^2=(-1.5)^2 = 2.25\), \((49 - 53.5)^2=(-4.5)^2 = 20.25\)

Step3: Calculate the variance

The variance \(s^{2}=\frac{600.25+72.25 + 272.25+272.25+812.25+42.25+2.25+20.25}{8 - 1}=\frac{2094}{7}\approx299.14\)

Step4: Calculate the standard deviation

The standard deviation \(s=\sqrt{299.14}\approx17.30\)

Step5: Add the new data point and calculate the new mean

The new data set is \(29,62,37,70,82,47,52,49,150\).
The new mean \(\bar{x}_{new}=\frac{29+62 + 37+70+82+47+52+49+150}{9}=\frac{578}{9}\approx64.22\)

Step6: Calculate the new squared - differences

\((29 - 64.22)^2=(-35.22)^2\approx1240.45\), \((62 - 64.22)^2=(-2.22)^2\approx4.93\), \((37 - 64.22)^2=(-27.22)^2\approx740.93\), \((70 - 64.22)^2=(5.78)^2\approx33.41\), \((82 - 64.22)^2=(17.78)^2\approx316.13\), \((47 - 64.22)^2=(-17.22)^2\approx296.53\), \((52 - 64.22)^2=(-12.22)^2\approx149.33\), \((49 - 64.22)^2=(-15.22)^2\approx231.65\), \((150 - 64.22)^2=(85.78)^2\approx7359.21\)

Step7: Calculate the new variance

The new variance \(s_{new}^{2}=\frac{1240.45+4.93+740.93+33.41+316.13+296.53+149.33+231.65+7359.21}{9 - 1}=\frac{10372.87}{8}=1296.61\)

Step8: Calculate the new standard deviation

The new standard deviation \(s_{new}=\sqrt{1296.61}\approx36.01\)
The mean increases from \(53.5\) to approximately \(64.22\) and the standard deviation increases from approximately \(17.30\) to approximately \(36.01\)

Answer:

a. Variance: approximately \(299.14\), Standard deviation: approximately \(17.30\)
b. The mean increases from approximately \(53.5\) to approximately \(64.22\) and the standard deviation increases from approximately \(17.30\) to approximately \(36.01\)