Question

a sample of 100 clients of an exercise facility was selected. let x = the number of days per week that a randomly selected client uses the exercise facility. find the number that is 1.5 standard deviations below the mean. (round your answer to three decimal places.)

Explanation:

Step1: Calculate the mean

The formula for the mean \(\mu=\sum_{i}x_{i}p_{i}\), where \(x_{i}\) is the value of the random variable and \(p_{i}=\frac{\text{Frequency}}{\text{Total}}\).
The total number of clients \(n = 100\).
\(\mu=(0\times\frac{2}{100})+(1\times\frac{15}{100})+(2\times\frac{30}{100})+(3\times\frac{29}{100})+(4\times\frac{10}{100})+(5\times\frac{9}{100})+(6\times\frac{5}{100})\)
\(\mu = 0 + 0.15+0.6 + 0.87+0.4+0.45 + 0.3\)
\(\mu=3.17\)

Step2: Calculate the variance

The formula for the variance \(\sigma^{2}=\sum_{i}(x_{i}-\mu)^{2}p_{i}\)
\((0 - 3.17)^{2}\times\frac{2}{100}+(1 - 3.17)^{2}\times\frac{15}{100}+(2 - 3.17)^{2}\times\frac{30}{100}+(3 - 3.17)^{2}\times\frac{29}{100}+(4 - 3.17)^{2}\times\frac{10}{100}+(5 - 3.17)^{2}\times\frac{9}{100}+(6 - 3.17)^{2}\times\frac{5}{100}\)
\(=(9.9969)\times\frac{2}{100}+(4.7089)\times\frac{15}{100}+(1.3689)\times\frac{30}{100}+(0.0289)\times\frac{29}{100}+(0.6889)\times\frac{10}{100}+(3.3489)\times\frac{9}{100}+(8.0089)\times\frac{5}{100}\)
\(=0.199938+0.706335+0.41067+0.008381+0.06889+0.301401+0.400445\)
\(\sigma^{2}=2.1\)

Step3: Calculate the standard deviation

\(\sigma=\sqrt{\sigma^{2}}=\sqrt{2.1}\approx1.449\)

Step4: Calculate the value 1.5 standard deviations below the mean

The formula is \(x=\mu - 1.5\sigma\)
\(x = 3.17-1.5\times1.449\)
\(x=3.17 - 2.1735\)
\(x = 0.9965\approx0.997\)

Answer:

\(0.997\)