Answer :
Answer:
Step-by-step explanation:
Let X be the monthly count of salmonellosis cases in South Dakota.
And X has Poisson(A 1.62)
E(X) = \lambda = 1.62
Var(X) = \lambda = 1.62
Standard Deviation (X) = \sqrt{\lambda} = \sqrt{1.62} = 1.27
c)
Mean of salmonellosis cases in South Dakota over a six month period is = 6*1.62 = 9.72
standard deviation of salmonellosis cases in South Dakota over a six month period is = 6*1.27 = 7.62
Answer:
A. Mean= 1.69
Standard deviation = 1.3
B. mean= 1.69
Standard deviation= 0.02
Step-by-step explanation:
A. FIND THE MEAN AND STANDARD DEVIATION;
Since 1.69 is the average number of salmonellosis cases per month in South Dakota. Therefore
Mean = 1.69
This mean is equal to the variance because we are not given a range of number to find the standard deviation.
Mean = variance
Standard deviation= √variance =√mean
Therefore;
Standard deviation = √1.69 = 1.3
B. FIND THE MEAN AND STANDARD DEVIATION OVER SIX MONTHS PERIOD;
Mean = 1.69
STANDARD DEVIATION:
Step 1: find the six values of the mean
Mean = sum of values÷ number of occurrence
1.69= sum of values / 6
Sum of values = 10.14
Let's assume the six numbers to be
1.66 + 1.67 + 1.68 + 1.70 + 1.71 + 1.72 = 10.14
Step2: subtract the mean from the value and square
(1.66 - 1.69)^2 = 0.0009
(1.67 - 1.69)^2= 0.0004
(1.68 - 1.69)^2 = 0.0001
(1.70 - 1.69)^2 = 0.0001
(1.71 - 1.69)^2 = 0.0004
(1.72 - 1.69)^2 = 0.0009
STEP3: add all the results in step 2
0.0009 + 0.0004 + 0.0001 + 0.0001 + 0.0004 + 0.0009 = 0.0028
STEP4: Find variance
0.0028 ÷ 6 = 0.0004667
STEP5 : Find standard deviation;
Standard deviation= √0.0004667 =
0.0216
To two significant figure, Therefore;
standard deviation= 0.02