Chapter 2, Problem 1RWYL

Raisins The number of raisins in each of 14miniboxes (1/2 -ounce size) was counted for ageneric brand and for Sunmaid brand raisins. Thetwo data sets are shown here:

a. What are the mean and the standard deviation for thegeneric brand?
b. What are the mean and the standard deviation for theSunmaid brand?
c. Compare the centers and variabilities of the twobrands using the results of parts a and b.

To determine

(a)

To find:

The mean and standard deviation for the generic brand.

Answer to Problem 1RWYL

The mean and standard deviation for the generic brand are $26.21$ and $1.2514$ respectively.

Explanation of Solution

Given information:

The given data is,

Generic Brand				Sunmaid
25	26	25	28	25	29	24	24
26	28	28	27	28	24	28	22
26	27	24	25	25	28	30	27
26	26			28	24

Calculation:

The data with calculation is shown in table below.

	$x$	${\left(x-\overline{x}\right)}^2$
	25	1.464 1
	26	0.044 1
	26	0.044 1
	26	0.044 1
	26	0.044 1
	28	3.204 1
	27	0.624 1
	26	0.044 1
	25	1.464 1
	28	3.204 1
	24	4.884 1
	28	3.204 1
	27	0.624 1
	25	1.464 1
Total	367	20.3574

  Table-1

The sample size is $n=14$.

The mean is,

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{367}{14}\\ =26.21\end{array}$

The standard deviation is,

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{20.3574}{14-1}}\\ =1.2514\end{array}$

Therefore, the mean and standard deviation for the generic brand are $26.21$ and $1.2514$ respectively.

To determine

(b)

To find:

The mean and standard deviation for Sunmaid.

Answer to Problem 1RWYL

The mean and standard deviation for the generic brand are $26.14$ and $2.4133$ respectively.

Explanation of Solution

Given information:

The given data is,

Generic Brand				Sunmaid
25	26	25	28	25	29	24	24
26	28	28	27	28	24	28	22
26	27	24	25	25	28	30	27
26	26			28	24

Calculation:

The data with calculation is shown in table below.

	$x$	${\left(x-\overline{x}\right)}^2$
	25	1.2996
	28	3.4596
	25	1.2996
	28	3.4596
	29	8.1796
	24	4.5796
	28	3.4596
	24	4.5796
	24	4.5796
	28	3.4596
	30	14.8996
	24	4.5796
	22	17.1396
	27	0.7396
Total	366	75.7144

  Table-2

The sample size is $n=14$.

The mean is,

  $\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x}}{n}\\ =\frac{366}{14}\\ =26.14\end{array}$

The standard deviation is,

  $\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n-1}}\\ =\sqrt{\frac{75.7144}{14-1}}\\ =2.4133\end{array}$

Therefore, the mean and standard deviation for the generic brand are $26.14$ and $2.4133$ respectively.

To determine

(c)

To compare:

The centers and variability of the brands.

Explanation of Solution

Given information:

The given data is,

Generic Brand				Sunmaid
25	26	25	28	25	29	24	24
26	28	28	27	28	24	28	22
26	27	24	25	25	28	30	27
26	26			28	24

The generic brand has smaller standard of deviation as compared to Sunmaid and the mean of generic brand is greater as compared to Sunmaid.