Question

a survey of 10 fast - food restaurants noted the number of calories in a mid - sized hamburger. the results are given in the table below.
calories in a mid - sized hamburger
514 506 502 498 497 507 459 477 464 513
find the mean and sample standard deviation of these data. round to the nearest hundredth.
mean
sample standard deviation

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}\) of a sample \(x_1,x_2,\cdots,x_n\) is \(\bar{x}=\frac{1}{n}\sum_{i = 1}^{n}x_i\).
Here \(n = 10\), and \(\sum_{i=1}^{10}x_i=514 + 506+502+498+497+507+459+477+464+513\)
\(\sum_{i=1}^{10}x_i = 4937\)
\(\bar{x}=\frac{4937}{10}=493.7\)

Step2: Calculate the sample standard deviation

The formula for the sample standard deviation \(s\) is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\)
First, calculate \((x_i-\bar{x})^2\) for each \(x_i\):
\((514 - 493.7)^2=(20.3)^2 = 412.09\)
\((506-493.7)^2=(12.3)^2 = 151.29\)
\((502 - 493.7)^2=(8.3)^2=68.89\)
\((498-493.7)^2=(4.3)^2 = 18.49\)
\((497-493.7)^2=(3.3)^2=10.89\)
\((507-493.7)^2=(13.3)^2 = 176.89\)
\((459-493.7)^2=(- 34.7)^2=1204.09\)
\((477-493.7)^2=(-16.7)^2 = 278.89\)
\((464-493.7)^2=(-29.7)^2=882.09\)
\((513-493.7)^2=(19.3)^2=372.49\)

\(\sum_{i = 1}^{10}(x_i-\bar{x})^2=412.09+151.29+68.89+18.49+10.89+176.89+1204.09+278.89+882.09+372.49\)
\(\sum_{i = 1}^{10}(x_i-\bar{x})^2=3576.1\)

Then \(s=\sqrt{\frac{3576.1}{10 - 1}}=\sqrt{\frac{3576.1}{9}}\approx\sqrt{397.3444}\approx19.93\)

Answer:

mean: \(493.70\)
sample standard deviation: \(19.93\)