Q. 3.211

Question


STD Surveillance. The Centers for Disease Control and Prevention compiles reported cases and rates of diseases in United States cities and outlying areas. In a document titled Sexually Transmitted Disease Surveillance, the number of reported cases of all stages of syphilis is provided for cities, including Orlando, Florida, and Cincinnati, Ohio. Following is the number of reported cases of syphilis for those two cities over the years 2007-2011.

a. Obtain the individual population means the number of cases for both cities.

b. Without doing any calculations, decide for which city the population standard deviation of the number of cases is smaller. Explain your answer.

c. Obtain the individual population standard deviations of the number of cases for both cities.

d. Are your answers to parts (b) and (c) consistent? Why or why not?

Step-by-Step Solution

Verified
Answer

(a) For Orlando mean is, μ=465.4 and for Cincinnati is, μ=265.8

(b) The city of Orlando has satisfied the given conditions.

(c) Standard deviation of Orlando is, σ=67.9311 and that of Cincinnati is, σ=167.0849

(d) Yes answer to part (c) and part (d) is consistent.

1Part (a) Step 1: Given information

We are given that,

Orlando,

The no. values of the data set are,N=5

And the data set is,

 xi=583, 460, 408, 391, 485

Cincinnati,

The no. values of the data set are, N=5

And the data set is,

xi=77, 105, 227, 484, 436

2Part (a) Step 2: Simplify

We know that, 

Mean is the sum of all values divided by number of values.

Orlando,

The mean is, 

   μ=xiN=583+460+408+391+4855=23275=465.4

Cincinnati,

The mean is,

  μ=xiN=77+105+227+484+4365=13295=265.8

3Part (b) Step 1: Given information

We need to find out the city whose standard deviation is the number of cases is smaller. 

4Part (b) Step 2: Explanation

We expect Orlando to have a smaller standard deviation than Cincinnati because its range is much smaller than the range of values for Cincinnati.

5Part (c) Step 1: Given information

We need to find out the standard deviation for both the cities.

6Part (c) Step 2: Simplify

As we know that standard deviation is the square root of sum of squared deviation divided by no. of values.

Orlando,

AGEDEVIATIONSQUARED DEVIATION
583
117.6
13829.76
460
-5.4
29.16
408
-57.4
3294.76
391
-74.4
5535.36
485
19.6
384.16
TOTAL23073.2

The population standard deviation is,

     σ=TotalN=23073.25=67.9311

Cincinnati,

AGEDEVIATIONSQUARED DEVIATION
77
-188.8
35645.44
105
-160.8
25856.64
227
-38.8
1505.44
484
218.2
47611.24
436
170.2
28968.04
TOTAL139586.8

The population standard deviation is,

σ=TotalN=139586.85=167.0849

7Part(d) Step 1: Given information

We need to find out whether parts (b) and (c) are consistent or not.

8Part (d) Step 2: Explanation

As we can interpret from parts (b) and (c). part (b) agrees with (c) because standard deviation measures the deviations of the mean and thus standard deviation is dependent upon range.