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

Expert verified

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

01

(a) Mean and standard deviation

Mean=120×96 + 93 +....+ 95 + 91X-=179120X-= 89.55S =1n-1i=1nx1-x¯2=119i=1nx1-2=4.71811

02

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

X-±S=84.8319,94.26811=1420×100=70%X-±2S=80.113,98.986Out of 20,19 numbers satisfied=1920×100=95%X-±3S=75.395,103.7043All numbers satisfied=100%

03

(b) Chebyshev’s

Chebyshev’s rule,

P1×x-μKσ1K2P1×x-μKσ1-1K2P1x-89.55151-1P1x-89.55150P1x-89.55251-14P1x-89.55250.75P1x-89.55351-19P1x-89.55350.889

Yes, it agrees with Chebyshev’s rule,

The empirical rule is,

x¯±σis 68%x¯±2σis 95%x¯±3σis 99.7%

The empirical rule is also satisfied.

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Most popular questions from this chapter

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