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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,

Leonid Glanz asked Jun 28, 2023

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Oliver Leung · Answer 1 · 2023-06-30T03:36:35+0000

The minimum, maximum, range, mean, and standard deviation of the given data set include;

Minimum = 17

Maximum = 91

Range = 74.

Mean = 53.04

Standard deviation = 23.67

In order to determine the five-number summary for the morning and afternoon dogs, we would re-arrange the data set in an ascending order (from least to greatest);

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

Based on the data set representing the number of hours spent watching TV, we can logically deduce that the five-number summary are as follows;

Minimum value = 17

Lower quartile = 29

Median = 52.5

Upper Quartile = 73.5

Maximum value: 91

In Mathematics and Statistics, the range of a data set can be calculated by using this mathematical expression;

Range = Highest number - Lowest number

Range = 91 - 17

Range = 74.

By using an online graphing calculator, the mean and standard deviation to two decimals are as follows;

Mean = 53.04

Standard deviation = 23.67.

The table below gives the number of hours spent watching TV last week by a sample-example-1

The table below gives the number of hours spent watching TV last week by a sample-example-2

Surfmuggle · Answer 2 · 2023-07-04T15:20:43+0000

ANSWER:

Min = 17

Max = 91

Range = 74

Mean = 53.04

Standard Deviation = 23.67

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:

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

Now we calculate the standard deviation:

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

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