Question

The data given below contain the state cigarette tax​ (in $) for all 20 regions of a country.

1.36
1.70
2.50
0.45
1.18
0.64
3.46
0.57
2.00
0.80
1.60
0.98
0.36
2.24
4.35
0.62
2.70
1.78
1.53
0.84

Required:
a. Compute the population mean and population standard deviation for the regional cigarette taxes.
b. The population standard deviation is σ=_______
c. Interpret the parameters in​ (a).

1. The population mean is the average amount of regional cigarette tax. The population standard deviation represents the amount of spread in the regional cigarette tax around the median.
2. The population mean is the most common regional cigarette tax. The population standard deviation represents the amount of spread in the regional cigarette tax around the median.
3. The population standard deviation is the average amount of regional cigarette tax. The population mean represents the amount of spread in the regional cigarette tax around the standard deviation.
4. The population mean is the average amount of regional cigarette tax. The population standard deviation represents the amount of spread in the regional cigarette tax around the mean.

Answer 1

(a)
`\bar x= 1.583` -- Population Mean

(b)
`s = 1.032` --- Population standard deviation

(c) See Explanation

Explanation:

Given:

Cigarette tax for 20 regions

Solving (a): The population mean

This is calculated as:

`\bar x = (\sum x)/(n)`

`\sum x = 1.36 + 1.70 + 2.50 + 0.45 + 1.18 + 0.64 + 3.46 + 0.57 + 2.00 + 0.80 + 1.60 +0.98 + 0.36 + 2.24 + 4.35 + 0.62 + 2.70 + 1.78 + 1.53 + 0.84`

`\sum x = 31.66`

`n = 20`

So, we have:

`\bar x= (31.66)/(20)`

`\bar x= 1.583`

Solving (b): The population standard deviation

This is calculated as:

`s = \sqrt{(\sum( x - \bar x)^2)/(n)`

`\sum (x -\bar x)^2 = (1.36 - 1.583)^2 + (1.70 - 1.583)^2+ (2.50 - 1.583)^2+ (0.45 - 1.583)^2+ (1.18 - 1.583)^2+ (0.64 - 1.583)^2+ (3.46 - 1.583)^2+ (0.57 - 1.583)^2+ (2.00 - 1.583)^2+ (0.80 - 1.583)^2+ (1.60 - 1.583)^2+(0.98 - 1.583)^2+ (0.36 - 1.583)^2+ (2.24 - 1.583)^2+ (4.35 - 1.583)^2+ (0.62 - 1.583)^2+ (2.70 - 1.583)^2+ (1.78 - 1.583)^2+ (1.53 - 1.583)^2+ (0.84- 1.583)^2`

`\sum (x -\bar x)^2 = 21.29222`

So:

`s = \sqrt{(21.2922)/(20)`

`s = √(1.06461)`

`s = 1.032`

Solving (c):

Population mean tells the average amount while the standard deviation represents the spread from the calculated mean

Option (4) is correct