Question

you are performing a study about weekly per capita milk consumption. a previous study found weekly per capita milk consumption to be normally distributed, with a mean of 47.7 fluid ounces and a standard deviation of 8.6 fluid ounces. you randomly sample 30 people and record the weekly milk consumptions shown below.
41 55 33 40 47 49 34 66 39 52
47 25 42 57 47 44 43 52 52 51
39 31 25 47 63 44 54 46 56 37
(a) draw a frequency histogram to display these data. use seven classes. choose the correct answer below.
(b) find the mean of your sample.
the mean is . (round to one decimal place as needed.)
(c) find the standard deviation of your sample.
the standard deviation is . (round to one decimal place as needed.)
(d) compare the mean and standard deviation of your sample with those of the previous study. discuss the differences.
the sample mean is than the previous mean, so, on average, consumption from the sample is than in the previous study. the sample standard deviation is than the previous standard deviation by , so the milk consumption is spread out in the sample.

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
We have \(n = 30\) data points: \(x=\{41,55,33,40,47,49,34,66,39,52,47,25,42,57,47,44,43,52,52,51,39,31,25,47,63,44,54,46,56,37\}\)
\(\sum_{i=1}^{30}x_{i}=41 + 55+33+40+47+49+34+66+39+52+47+25+42+57+47+44+43+52+52+51+39+31+25+47+63+44+54+46+56+37=1410\)
\(\bar{x}=\frac{1410}{30}=47\)

Step2: Calculate the variance

The formula for the variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
\((x_{1}-\bar{x})^{2}=(41 - 47)^{2}=(-6)^{2}=36\)
\((x_{2}-\bar{x})^{2}=(55 - 47)^{2}=8^{2}=64\)
\((x_{3}-\bar{x})^{2}=(33 - 47)^{2}=(-14)^{2}=196\)
\((x_{4}-\bar{x})^{2}=(40 - 47)^{2}=(-7)^{2}=49\)
\((x_{5}-\bar{x})^{2}=(47 - 47)^{2}=0^{2}=0\)
\((x_{6}-\bar{x})^{2}=(49 - 47)^{2}=2^{2}=4\)
\((x_{7}-\bar{x})^{2}=(34 - 47)^{2}=(-13)^{2}=169\)
\((x_{8}-\bar{x})^{2}=(66 - 47)^{2}=19^{2}=361\)
\((x_{9}-\bar{x})^{2}=(39 - 47)^{2}=(-8)^{2}=64\)
\((x_{10}-\bar{x})^{2}=(52 - 47)^{2}=5^{2}=25\)
\((x_{11}-\bar{x})^{2}=(47 - 47)^{2}=0^{2}=0\)
\((x_{12}-\bar{x})^{2}=(25 - 47)^{2}=(-22)^{2}=484\)
\((x_{13}-\bar{x})^{2}=(42 - 47)^{2}=(-5)^{2}=25\)
\((x_{14}-\bar{x})^{2}=(57 - 47)^{2}=10^{2}=100\)
\((x_{15}-\bar{x})^{2}=(47 - 47)^{2}=0^{2}=0\)
\((x_{16}-\bar{x})^{2}=(44 - 47)^{2}=(-3)^{2}=9\)
\((x_{17}-\bar{x})^{2}=(43 - 47)^{2}=(-4)^{2}=16\)
\((x_{18}-\bar{x})^{2}=(52 - 47)^{2}=5^{2}=25\)
\((x_{19}-\bar{x})^{2}=(52 - 47)^{2}=5^{2}=25\)
\((x_{20}-\bar{x})^{2}=(51 - 47)^{2}=4^{2}=16\)
\((x_{21}-\bar{x})^{2}=(39 - 47)^{2}=(-8)^{2}=64\)
\((x_{22}-\bar{x})^{2}=(31 - 47)^{2}=(-16)^{2}=256\)
\((x_{23}-\bar{x})^{2}=(25 - 47)^{2}=(-22)^{2}=484\)
\((x_{24}-\bar{x})^{2}=(47 - 47)^{2}=0^{2}=0\)
\((x_{25}-\bar{x})^{2}=(63 - 47)^{2}=16^{2}=256\)
\((x_{26}-\bar{x})^{2}=(44 - 47)^{2}=(-3)^{2}=9\)
\((x_{27}-\bar{x})^{2}=(54 - 47)^{2}=7^{2}=49\)
\((x_{28}-\bar{x})^{2}=(46 - 47)^{2}=(-1)^{2}=1\)
\((x_{29}-\bar{x})^{2}=(56 - 47)^{2}=9^{2}=81\)
\((x_{30}-\bar{x})^{2}=(37 - 47)^{2}=(-10)^{2}=100\)

\(\sum_{i = 1}^{30}(x_{i}-\bar{x})^{2}=36+64 + 196+49+0+4+169+361+64+25+0+484+25+100+0+9+16+25+25+16+64+256+484+0+256+9+49+1+81+100=3690\)

\(s^{2}=\frac{3690}{29}\approx127.24\)

Step3: Calculate the standard deviation

The formula for the standard deviation \(s=\sqrt{s^{2}}\)
\(s=\sqrt{127.24}\approx11.3\)

Answer:

The mean is \(47\). The standard deviation is \(11.3\)