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.
yes, the frequency histogram of the data is bell
shaped.
no, the frequency histogram of the data is not 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 mid - points

The mid - points \(x_i\) for each class:
For \(50 - 59\): \(x_1=\frac{50 + 59}{2}=54.5\)
For \(60 - 69\): \(x_2=\frac{60+69}{2}=64.5\)
For \(70 - 79\): \(x_3=\frac{70 + 79}{2}=74.5\)
For \(80 - 89\): \(x_4=\frac{80+89}{2}=84.5\)
For \(90 - 99\): \(x_5=\frac{90 + 99}{2}=94.5\)
For \(100 - 109\): \(x_6=\frac{100+109}{2}=104.5\)

Step2: Calculate the mean \(\mu\)

The formula for the mean of a frequency distribution is \(\mu=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\)
\(\sum_{i = 1}^{n}f_ix_i=2\times54.5+316\times64.5+1458\times74.5+1517\times84.5+228\times94.5+13\times104.5\)
\(=109+20382+108621+128131.5+21546+1358.5\)
\(=280148\)
\(\sum_{i=1}^{n}f_i=2 + 316+1458+1517+228+13=3534\)
\(\mu=\frac{280148}{3534}\approx79.3\)

Step3: Calculate the variance \(\sigma^{2}\)

The formula for the variance is \(\sigma^{2}=\frac{\sum_{i = 1}^{n}f_i(x_i-\mu)^{2}}{\sum_{i=1}^{n}f_i}\)
\((x_1-\mu)^{2}=(54.5 - 79.3)^{2}=(-24.8)^{2}=615.04\)
\((x_2-\mu)^{2}=(64.5 - 79.3)^{2}=(-14.8)^{2}=219.04\)
\((x_3-\mu)^{2}=(74.5 - 79.3)^{2}=(-4.8)^{2}=23.04\)
\((x_4-\mu)^{2}=(84.5 - 79.3)^{2}=(5.2)^{2}=27.04\)
\((x_5-\mu)^{2}=(94.5 - 79.3)^{2}=(15.2)^{2}=231.04\)
\((x_6-\mu)^{2}=(104.5 - 79.3)^{2}=(25.2)^{2}=635.04\)

\(\sum_{i = 1}^{n}f_i(x_i-\mu)^{2}=2\times615.04+316\times219.04+1458\times23.04+1517\times27.04+228\times231.04+13\times635.04\)
\(=1230.08+69216.64+33692.32+41029.68+52677.12+8255.52\)
\(=206101.36\)

\(\sigma^{2}=\frac{206101.36}{3534}\approx58.3\)

Step4: Calculate the standard deviation \(\sigma\)

\(\sigma=\sqrt{\sigma^{2}}=\sqrt{58.3}\approx7.6\)

Step5: Use the empirical rule for part (c)

The empirical rule states that for a bell - shaped distribution, about \(95\%\) of the data lies within \(\mu\pm2\sigma\)
\(\mu - 2\sigma=79.3-2\times7.6=79.3 - 15.2 = 64.1\)
\(\mu+2\sigma=79.3 + 2\times7.6=79.3+15.2 = 94.5\)

Answer:

(a) \(\mu\approx79.3\), \(\sigma\approx7.6\)
(b) 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 \(64.1\) and \(94.5\)