In the decade 1982 through 1991, 10 employees working at the Amoco Company chemi
ID: 3135015 • Letter: I
Question
In the decade 1982 through 1991, 10 employees working at the Amoco Company chemical research center were stricken with brain tumors. The average employment at the center was 2000 employees. Nationwide, the average incidence of brain tumors in a single year is 20 per 100,000 people. If the incidence of brain tumors at the Amoco chemical research center were the same as the nationwide incidence, what is the probability that at least 10 brain tumors would have been observed among Amoco workers during the decade 1982 through 1991? What do you conclude from your analysis? (Source: AP wire service report, March 12, 1994)
Explanation / Answer
WE HAVE THE TOTAL TRY = 2000 = N
THE PROBABILITY OF TUMORS = 20 PER 100000
= 20/100000 = 0.0002
NOW WE HAVE TO FIND PROBABILITY OF GETTING BRAIN TUMORS TO ATLEAST 10 OUT OF 2000
WHICH WILL BE = P(X>=10) = 1 - [P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+]
AND EACH P CAN BE CALCULATED IN A MANNER AS BELOW
P(X=0) = 2000C0*(0.0002)^0*(0.9998)^2000 = 0.670
SIMILARLY WE HAVE TO CALCULATE IT TILL P(X=9)
AND WE WILL ADD THERE PROBABILITY
AND TO GET THR P(X>=10) WE WILL SUBTRACT THE SUM FROM 1
THE ANSWER CALCULATED FROM THE SCIENTIFIC CALCULATOR CAME OUT TO BE = 0.0806
HENCE P(X>=10) = 0.0806