Question

Sanitation inspection of cruise ships. Refer to the Centres for Disease Control and Prevention listing of the October 2015 sanitation scores for 20 cruise ships, Exercise 2.24 (p. 83).

a. Find the mean and standard deviation of the sanitation scores.

b. Calculate the intervals x {s, x {2s, x {3s.

c. Find the percentage of measurements in the data set that fall within each interval, part

b. Do these percentages agree with Chebyshev’s Rule? The Empirical Rule?

Short answer

a) Mean = 89.55

S = 4.71811

b) The interval for s (84.8319, 94.26811)

The interval for 2s (80.113, 98.986)

The interval for 3s (75.395, 103.7043)

c) s is 70%

2s is 95%

3s is 100%

b) Both are agreeable.

Step-by-step solution

(a) Mean and standard deviation

$$\begin{gathered} Mean=\frac{1}{20}\times \left(\text{96 + 93 +}....\text{+ 95 + 91}\right) \\ \stackrel{-}{\text{X}}\text{=}\frac{\text{1791}}{\text{20}} \\ \stackrel{-}{\text{X}}\text{= 89.55} \\ \text{S =}\sqrt{\frac{1}{n-1}}\sum _{i=1}^n{\left(x_1-\overline{x}\right)}^2 \\ =\sqrt{\frac{1}{19}}\sum _{i=1}^n{\left(x_1-\right)}^2 \\ =4.71811 \end{gathered}$$

(b) or (c) Intervals calculate and percentage of measurement

$$\begin{gathered} \stackrel{-}{\text{X}}\pm S=\left(84.8319,94.26811\right) \\ =\frac{\text{14}}{20}\times 100 \\ =70\% \\ \stackrel{-}{\text{X}}\pm 2S=\left(80.113,98.986\right) \\ \text{Out of 20,19 numbers satisfied} \\ =\frac{\text{19}}{20}\times 100 \\ =95\% \\ \stackrel{-}{\text{X}}\pm 3S\text{=}\left(75.395,103.7043\right) \\ \text{All numbers satisfied} \\ =100\% \end{gathered}$$

(b) Chebyshev’s

Chebyshev’s rule,

$$\begin{gathered} \text{P}\left(1\times x-\left.\mu \right|⩾K\sigma \right)⩽\frac{1}{K^2} \\ \text{P}\left(1\times x-\left.\mu \right|⩾K\sigma \right)⩾1-\frac{1}{K^2} \\ \text{P}\left(1x-\left.89.55\right|⩽15\right)⩾1-1 \\ \text{P}\left(1x-\left.89.55\right|⩽15\right)⩾0 \\ \text{P}\left(1x-\left.89.55\right|⩽25\right)⩾1-\frac{1}{4} \\ \text{P}\left(1x-\left.89.55\right|⩽25\right)⩾0.75 \\ \text{P}\left(1x-\left.89.55\right|⩽35\right)⩾1-\frac{1}{9} \\ \text{P}\left(1x-\left.89.55\right|⩽35\right)⩾0.889 \end{gathered}$$

Yes, it agrees with Chebyshev’s rule,

The empirical rule is,

$$\begin{gathered} \overline{x}\pm \sigma \text{is 68\%} \\ \overline{x}\pm 2\sigma \text{is 95\%} \\ \overline{x}\pm 3\sigma \text{is 99.7\%} \end{gathered}$$

The empirical rule is also satisfied.