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a random sample of 53 students was selected, and they were asked the nu…

Question

a random sample of 53 students was selected, and they were asked the number of pairs of shoes they. the results are as follows:

# of pairs of shoes45678910111213
frequency10366296335

round answers to 4 decimal places.
the mean, $\bar{x}$, is:
the sample standard deviation, $s$, is:

Explanation:

Step1: Calculate the product of each value and its frequency

Let \(x_i\) be the number of pairs of shoes and \(f_i\) be the frequency. The sum \(\sum_{i = 1}^{n}x_if_i=4\times10 + 5\times3+6\times6 + 7\times6+8\times2+9\times9+10\times6+11\times3+12\times3+13\times5\)
\[

$$\begin{align*} &4\times10+5\times3 + 6\times6+7\times6+8\times2+9\times9+10\times6+11\times3+12\times3+13\times5\\ =&40+15 + 36+42+16+81+60+33+36+65\\ =&424 \end{align*}$$

\]
The total frequency \(N=\sum_{i = 1}^{n}f_i=10 + 3+6+6+2+9+6+3+3+5=53\)

Step2: Calculate the mean

The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{N}=\frac{424}{53}=8\)

Step3: Calculate the squared - differences and their sum

The formula for the sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^{2}}{N - 1}\)
\[

$$\begin{align*} &\sum_{i = 1}^{n}f_i(x_i - \bar{x})^{2}=10\times(4 - 8)^{2}+3\times(5 - 8)^{2}+6\times(6 - 8)^{2}+6\times(7 - 8)^{2}+2\times(8 - 8)^{2}+9\times(9 - 8)^{2}+6\times(10 - 8)^{2}+3\times(11 - 8)^{2}+3\times(12 - 8)^{2}+5\times(13 - 8)^{2}\\ =&10\times16+3\times9+6\times4+6\times1+2\times0+9\times1+6\times4+3\times9+3\times16+5\times25\\ =&160 + 27+24+6+0+9+24+27+48+125\\ =&450 \end{align*}$$

\]

Step4: Calculate the sample variance

The sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^{2}}{N - 1}=\frac{450}{53 - 1}=\frac{450}{52}\approx8.6538\)

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^{2}}=\sqrt{\frac{450}{52}}\approx2.9417\)

Answer:

The mean, \(\bar{x}\), is \(8.0000\)
The sample standard deviation, \(s\), is \(2.9417\)