Question

question 11
the table below gives the number of hours spent watching tv last week by a sample of 24 child
39 55 63 20 67 71
52 89 79 25 68 55
22 20 36 24 69 54
35 89 70 34 93 17
a. range =
b. standard deviation =
round your answers to the nearest hundredth as needed.
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Explanation:

Step1: Calculate the range

The range is the difference between the maximum and minimum values in the data set.
First, find the maximum value: \(93\)
Then, find the minimum value: \(17\)
Range \(=93 - 17=76\)

Step2: Calculate the standard deviation

1. Calculate the mean \(\bar{x}\)

The data set \(x=\{39,55,63,20,67,71,52,89,79,25,68,55,22,20,36,24,69,54,35,89,70,34,93,17\}\)
\(n = 24\)
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{39+55+63+20+67+71+52+89+79+25+68+55+22+20+36+24+69+54+35+89+70+34+93+17}{24}\)
\(\sum_{i=1}^{24}x_{i}=39 + 55+63+20+67+71+52+89+79+25+68+55+22+20+36+24+69+54+35+89+70+34+93+17 = 1330\)
\(\bar{x}=\frac{1330}{24}\approx55.42\)

1. Calculate the variance \(s^{2}\)

\(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\((x_{1}-\bar{x})^{2}=(39 - 55.42)^{2}=(- 16.42)^{2}=269.6164\)
\((x_{2}-\bar{x})^{2}=(55 - 55.42)^{2}=(-0.42)^{2}=0.1764\)
\((x_{3}-\bar{x})^{2}=(63 - 55.42)^{2}=(7.58)^{2}=57.4564\)
\((x_{4}-\bar{x})^{2}=(20 - 55.42)^{2}=(-35.42)^{2}=1254.5764\)
\((x_{5}-\bar{x})^{2}=(67 - 55.42)^{2}=(11.58)^{2}=134.1764\)
\((x_{6}-\bar{x})^{2}=(71 - 55.42)^{2}=(15.58)^{2}=242.7364\)
\((x_{7}-\bar{x})^{2}=(52 - 55.42)^{2}=(-3.42)^{2}=11.6964\)
\((x_{8}-\bar{x})^{2}=(89 - 55.42)^{2}=(33.58)^{2}=1127.6164\)
\((x_{9}-\bar{x})^{2}=(79 - 55.42)^{2}=(23.58)^{2}=556.0164\)
\((x_{10}-\bar{x})^{2}=(25 - 55.42)^{2}=(-30.42)^{2}=925.3764\)
\((x_{11}-\bar{x})^{2}=(68 - 55.42)^{2}=(12.58)^{2}=158.2564\)
\((x_{12}-\bar{x})^{2}=(55 - 55.42)^{2}=(-0.42)^{2}=0.1764\)
\((x_{13}-\bar{x})^{2}=(22 - 55.42)^{2}=(-33.42)^{2}=1116.8964\)
\((x_{14}-\bar{x})^{2}=(20 - 55.42)^{2}=(-35.42)^{2}=1254.5764\)
\((x_{15}-\bar{x})^{2}=(36 - 55.42)^{2}=(-19.42)^{2}=377.1364\)
\((x_{16}-\bar{x})^{2}=(24 - 55.42)^{2}=(-31.42)^{2}=987.2164\)
\((x_{17}-\bar{x})^{2}=(69 - 55.42)^{2}=(13.58)^{2}=184.4164\)
\((x_{18}-\bar{x})^{2}=(54 - 55.42)^{2}=(-1.42)^{2}=2.0164\)
\((x_{19}-\bar{x})^{2}=(35 - 55.42)^{2}=(-20.42)^{2}=416.9764\)
\((x_{20}-\bar{x})^{2}=(89 - 55.42)^{2}=(33.58)^{2}=1127.6164\)
\((x_{21}-\bar{x})^{2}=(70 - 55.42)^{2}=(14.58)^{2}=212.5764\)
\((x_{22}-\bar{x})^{2}=(34 - 55.42)^{2}=(-21.42)^{2}=458.8164\)
\((x_{23}-\bar{x})^{2}=(93 - 55.42)^{2}=(37.58)^{2}=1412.2564\)
\((x_{24}-\bar{x})^{2}=(17 - 55.42)^{2}=(-38.42)^{2}=1476.0964\)

\(\sum_{i = 1}^{24}(x_{i}-\bar{x})^{2}=269.6164+0.1764 + 57.4564+1254.5764+134.1764+242.7364+11.6964+1127.6164+556.0164+925.3764+158.2564+0.1764+1116.8964+1254.5764+377.1364+987.2164+184.4164+2.0164+416.9764+1127.6164+212.5764+458.8164+1412.2564+1476.0964=15993.33\)

\(s^{2}=\frac{15993.33}{23}\approx695.36\)

1. Calculate the standard deviation \(s\)

\(s=\sqrt{s^{2}}=\sqrt{695.36}\approx26.37\)

Answer:

a. \(76\)
b. \(26.37\)