Question

The table below gives the number of hours spent watching TV last week by a sample of 24 children. Find the minimum, maximum, range, mean, and standard deviation of the following data using using Excel, a calculator or other technology. The minimum, maximum, range should be entered as exact values. Round the mean and standard deviation to two decimals places.26 29 60 28 37 7268 29 65 85 74 6888 26 59 46 40 8031 45 86 23 91 17Min =Max =RangeMean =Standard Deviation =

Answer 1

ANSWER:

Min = 17

Max = 91

Range = 74

Mean = 53.04

Standard Deviation = 23.67

STEP-BY-STEP EXPLANATION:

We have the following data set:

$26,\:29,\:60,\:28,\:37,\:72,\:68,\:29,\:65,\:85,\:74,\:68,\:88,\:26,\:59,\:46,\:40,\:80,\:31,\:45,\:86,\:23,\:91,\:17$

The first thing we need to do is organize the data, as follows (ascending order):

$17,\:23,\:26,\:26,\:28,\:29,\:29,\:31,\:37,\:40,\:45,\:46,\:59,\:60,\:65,\:68,\:68,\:72,\:74,\:80,\:85,\:86,\:88,\:91$

In this way we can determine the minimum, maximum value and the range:

$\begin{gathered} Min=17 \\ \\ Max=91 \\ \\ Range=91-17=74 \end{gathered}$

We calculate the mean as follows:

$\begin{gathered} \mu=\frac{26+29+60+28+37+72+68+29+65+85+74+68+88+26+59+46+40+80+31+45+86+23+91+17}{24} \\ \\ \mu=\frac{1273}{24} \\ \\ \mu=53.04 \end{gathered}$

Now we calculate the standard deviation:

$\begin{gathered} \sigma=\sqrt{\frac{1}{n}\sum^n(x_i-\mu)^2} \\ \\ (x_i-\mu)^2=\lparen26-53.04)^2+\left(29-53.04\right)^2+\lparen60-53.04)^2+\lparen28-53.04)^2+\lparen37-53.04)^2+\left(72-53.04\right)^2+\lparen68-53.04)^2+\lparen29-53.04)^2+\lparen65-53.04)^2+\left(85-53.04\right)^2+\lparen74-53.04)^2+\lparen68-53.04)^2+\lparen88-53.04)^2+\left(26-53.04\right)^2+\lparen59-53.04)^2+\lparen46-53.04)^2+\lparen40-53.04)^2+\left(80-53.04\right)^2+\lparen31-53.04)^2+\lparen45-53.04)^2+\lparen86-53.04)^2+\left(23-53.04\right)^2 \\ \\ (x_i-\mu)^2=13444.96 \\ \\ \frac{1}{n}\operatorname{\sum}^n(x_i-\mu)^2=\frac{13444.95}{24}=560.21 \\ \\ \sigma=\sqrt{560.21} \\ \\ \sigma=23.67 \end{gathered}$