Question

the following data represent the high - temperature distribution for a summer month in a city for some of the last 130 years. treat the data as a population. complete parts (a) through (c).
(a) approximate the mean and standard deviation for temperature.
\\( \mu=\square \\) (round to one decimal place as needed.)
\\( \sigma=\square \\) (round to one decimal place as needed.)
(b) use the frequency histogram of the data to verify that the distribution is bell - shaped.
no, the frequency histogram of the data is not bell
shaped.
yes, the frequency histogram of the data is bell
shaped.
(c) according to the empirical rule, \\( 95 \\% \\) of days in the month will be between what two temperatures?
\\( \square \\) and \\( \square \\) (round to one decimal place as needed. use ascending order.)

Explanation:

Step1: Calculate the mean ($\mu$)

The formula for the mean of a grouped data is $\mu=\frac{\sum_{i = 1}^{n}f_ix_i}{N}$, where $f_i$ is the frequency and $x_i$ is the mid - point of the class interval.
For the class interval \(50 - 59\), \(x_1 = 54.5\), \(f_1=2\)
For the class interval \(60 - 69\), \(x_2 = 64.5\), \(f_2 = 307\)
For the class interval \(70 - 79\), \(x_3 = 74.5\), \(f_3=1454\)
For the class interval \(80 - 89\), \(x_4 = 84.5\), \(f_4=1486\)
For the class interval \(90 - 99\), \(x_5 = 94.5\), \(f_5=459\)
For the class interval \(100 - 109\), \(x_6 = 104.5\), \(f_6=12\)
\(N=\sum_{i = 1}^{6}f_i=2 + 307+1454+1486+459+12=3720\)
\(\sum_{i = 1}^{6}f_ix_i=2\times54.5+307\times64.5+1454\times74.5+1486\times84.5+459\times94.5+12\times104.5\)
\(=109+19801.5+108323+125567+43375.5+1254\)
\(=109+(19801.5+108323)+(125567+43375.5)+1254\)
\(=109 + 128124.5+168942.5+1254\)
\(=(109+128124.5)+(168942.5+1254)\)
\(=128233.5+170196.5\)
\(=298430\)
\(\mu=\frac{298430}{3720}\approx80.2\)

Step2: Calculate the standard deviation ($\sigma$)

The formula for the standard deviation of a grouped data is \(\sigma=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\mu)^2}{N}}\)
\((x_1-\mu)^2=(54.5 - 80.2)^2=(- 25.7)^2 = 660.49\), \(f_1(x_1-\mu)^2=2\times660.49 = 1320.98\)
\((x_2-\mu)^2=(64.5 - 80.2)^2=(-15.7)^2 = 246.49\), \(f_2(x_2-\mu)^2=307\times246.49=75672.43\)
\((x_3-\mu)^2=(74.5 - 80.2)^2=(-5.7)^2 = 32.49\), \(f_3(x_3-\mu)^2=1454\times32.49 = 47230.46\)
\((x_4-\mu)^2=(84.5 - 80.2)^2=(4.3)^2 = 18.49\), \(f_4(x_4-\mu)^2=1486\times18.49=27476.14\)
\((x_5-\mu)^2=(94.5 - 80.2)^2=(14.3)^2 = 204.49\), \(f_5(x_5-\mu)^2=459\times204.49 = 93760.91\)
\((x_6-\mu)^2=(104.5 - 80.2)^2=(24.3)^2 = 590.49\), \(f_6(x_6-\mu)^2=12\times590.49 = 7085.88\)
\(\sum_{i = 1}^{6}f_i(x_i-\mu)^2=1320.98+75672.43+47230.46+27476.14+93760.91+7085.88\)
\(=1320.98+(75672.43+47230.46)+(27476.14+93760.91)+7085.88\)
\(=1320.98 + 122902.89+121237.05+7085.88\)
\(=(1320.98+122902.89)+(121237.05+7085.88)\)
\(=124223.87+128322.93\)
\(=252546.8\)
\(\sigma=\sqrt{\frac{252546.8}{3720}}\approx\sqrt{67.89}\approx8.2\)

Step3: Check the shape of the histogram (part b)

Visually, from the histogram (description), the distribution is bell - shaped.

Step4: Use the empirical rule (part c)

According to the empirical rule, for a bell - shaped (normal) distribution, about \(95\%\) of the data lies within \(\mu\pm2\sigma\)
\(\mu - 2\sigma=80.2-2\times8.2=80.2 - 16.4 = 63.8\)
\(\mu + 2\sigma=80.2+2\times8.2=80.2 + 16.4 = 96.6\)

Answer:

(a) \(\mu = 80.2\), \(\sigma=8.2\)
(b) Yes, the frequency histogram of the data is bell - shaped.
(c) \(95\%\) of days in the month will be between \(63.8\) and \(96.6\)