Question

the results of a random sample of the number of pets per household in a region are shown in the histogram. estimate the sample mean and the sample standard deviation of the data set. the sample mean is $\bar{x}=2.0$ (round to one decimal place as needed.) the sample standard deviation is $s=square$ (round to one decimal place as needed.)

Explanation:

Step1: Identify mid - points and frequencies

Let the number of pets be grouped as \(0 - 1\), \(1 - 2\), \(2 - 3\), \(3 - 4\). Mid - points \(x_i\) are \(0.5\), \(1.5\), \(2.5\), \(3.5\) and frequencies \(f_i\) are \(4\), \(12\), \(8\), \(6\) respectively.

Step2: Calculate the sample mean \(\bar{x}\)

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\).
\(\sum_{i = 1}^{n}f_ix_i=4\times0.5 + 12\times1.5+8\times2.5 + 6\times3.5=2+18 + 20+21 = 61\).
\(\sum_{i=1}^{n}f_i=4 + 12+8 + 6=30\).
\(\bar{x}=\frac{61}{30}\approx2.0\) (already given).

Step3: Calculate \(\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2\)

\((x_1-\bar{x})^2=(0.5 - 2.0)^2=(-1.5)^2 = 2.25\), \(f_1(x_1-\bar{x})^2=4\times2.25 = 9\).
\((x_2-\bar{x})^2=(1.5 - 2.0)^2=(-0.5)^2 = 0.25\), \(f_2(x_2-\bar{x})^2=12\times0.25 = 3\).
\((x_3-\bar{x})^2=(2.5 - 2.0)^2=(0.5)^2 = 0.25\), \(f_3(x_3-\bar{x})^2=8\times0.25 = 2\).
\((x_4-\bar{x})^2=(3.5 - 2.0)^2=(1.5)^2 = 2.25\), \(f_4(x_4-\bar{x})^2=6\times2.25 = 13.5\).
\(\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=9 + 3+2+13.5 = 27.5\).

Step4: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}\), where \(n=\sum_{i=1}^{n}f_i = 30\).
\(s=\sqrt{\frac{27.5}{30 - 1}}=\sqrt{\frac{27.5}{29}}\approx0.97\approx1.0\).

Answer:

The sample standard deviation is \(s = 1.0\)