Question

use the appropriate formulas and methods to answer the following questions. do not use technology as an aid, unless otherwise stated. find the range and standard deviation for the data items in the histogram below. round the standard deviation to two decimal places if needed. the range is. the standard deviation is. hint

Explanation:

Step1: Find the range

The range is the difference between the maximum and minimum data - values. From the histogram, the minimum data - value is 10 and the maximum data - value is 70.
$Range=70 - 10$

Step2: Calculate the mean

First, find the mid - points of each class interval. The intervals are 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70 with mid - points $x_1 = 15$, $x_2=25$, $x_3 = 35$, $x_4 = 45$, $x_5 = 55$, $x_6 = 65$ and frequencies $f_1 = 1$, $f_2 = 2$, $f_3 = 1$, $f_4 = 6$, $f_5 = 2$, $f_6 = 2$.
The total frequency $n=\sum_{i = 1}^{6}f_i=1 + 2+1 + 6+2 + 2=14$.
The mean $\bar{x}=\frac{\sum_{i = 1}^{6}f_ix_i}{n}=\frac{1\times15+2\times25 + 1\times35+6\times45+2\times55+2\times65}{14}=\frac{15 + 50+35 + 270+110+130}{14}=\frac{610}{14}\approx43.57$.

Step3: Calculate the squared - deviations

Calculate $(x_i-\bar{x})^2$ for each $i$ and multiply by the frequency $f_i$.
For $i = 1$: $f_1(x_1-\bar{x})^2=1\times(15 - 43.57)^2=1\times(- 28.57)^2=1\times816.2449 = 816.2449$.
For $i = 2$: $f_2(x_2-\bar{x})^2=2\times(25 - 43.57)^2=2\times(-18.57)^2=2\times344.8449 = 689.6898$.
For $i = 3$: $f_3(x_3-\bar{x})^2=1\times(35 - 43.57)^2=1\times(-8.57)^2=1\times73.4449 = 73.4449$.
For $i = 4$: $f_4(x_4-\bar{x})^2=6\times(45 - 43.57)^2=6\times(1.43)^2=6\times2.0449 = 12.2694$.
For $i = 5$: $f_5(x_5-\bar{x})^2=2\times(55 - 43.57)^2=2\times(11.43)^2=2\times130.6449 = 261.2898$.
For $i = 6$: $f_6(x_6-\bar{x})^2=2\times(65 - 43.57)^2=2\times(21.43)^2=2\times459.2449 = 918.4898$.
The sum of squared - deviations $\sum_{i = 1}^{6}f_i(x_i-\bar{x})^2=816.2449+689.6898+73.4449+12.2694+261.2898+918.4898 = 2771.4386$.

Step4: Calculate the standard deviation

The standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{6}f_i(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{2771.4386}{14 - 1}}=\sqrt{\frac{2771.4386}{13}}\approx\sqrt{213.1876}\approx14.60$.

Answer:

The range is 60.
The standard deviation is 14.60.