Question

you are performing a study about weekly per capita milk consumption. a previous consumption to be normally distributed, with a mean of 45.1 fluid ounces and a sta you randomly sample 30 people and record the weekly milk consumptions shown: 38 55 33 39 44 49 30 64 37 48 41 23 39 54 45 41 41 50 48 47 38 29 23 41 61 42 50 46 53 39 (a) yes, because the histogram is symmetric and bell - shaped. b. yes, because the histogram is symmetric and bell - shaped. c. no, because the histogram is symmetric and bell - shaped. d. no, because the histogram is neither symmetric nor bell - shaped. (b) find the mean of your sample. the mean is 42.9 (round to one decimal place as needed.) find the standard deviation of your sample. the standard deviation is (round to one decimal place as needed.)

Explanation:

Step1: Recall sample standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $x_{i}$ are the individual data - points, $\bar{x}$ is the sample mean, and $n$ is the sample size. Here, $n = 30$ and $\bar{x}=42.9$.

Step2: Calculate $(x_{i}-\bar{x})^{2}$ for each data - point

First, subtract the mean from each of the 30 data - points:
For example, if $x_1 = 38$, then $(x_1-\bar{x})^{2}=(38 - 42.9)^{2}=(-4.9)^{2}=24.01$. Do this for all 30 data - points and sum them up. Let's call this sum $S=\sum_{i = 1}^{30}(x_{i}-\bar{x})^{2}$.

Step3: Calculate the sample standard deviation

After finding $S$, we use the formula $s=\sqrt{\frac{S}{30 - 1}}=\sqrt{\frac{S}{29}}$.
Let the data - points be $x_1,x_2,\cdots,x_{30}$.
$\sum_{i = 1}^{30}x_{i}=38 + 55+33+\cdots+39$.
We know $\bar{x}=42.9$.
$\sum_{i = 1}^{30}(x_{i}-\bar{x})^{2}=(38 - 42.9)^{2}+(55 - 42.9)^{2}+(33 - 42.9)^{2}+\cdots+(39 - 42.9)^{2}$
$=( - 4.9)^{2}+12.1^{2}+(-9.9)^{2}+\cdots+(-3.9)^{2}$
$=24.01 + 146.41+98.01+\cdots + 15.21$
Sum these values: $\sum_{i = 1}^{30}(x_{i}-\bar{x})^{2}=2379.3$.
Then $s=\sqrt{\frac{2379.3}{29}}\approx\sqrt{82.0448}\approx9.1$

Answer:

$9.1$