Question
In: Statistics and Probability
Assume that a procedure yields a binomial distribution with n trials and the probability of success...
Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma. ?Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. n
n=1510?, p=4/ 5
Solutions
Expert Solution
Solution :
Given that,
n = 1510
p = 4/5 = 0.80
q = 1 - p = 1 - 0.80 = 0.20
mean = 
= n * p = 1510 * 0.80 = 1208
standard deviation = 
= 
n * p * q =
1510 * 0.80 * 0.20 = 15.54
Minimum usual value =
- 2
= 1208 - 2 * 15.54 = 1176.92
Maximum usual value =
+ 2
= 1208 + 2 * 15.54 = 1239.08
orchestra answered 2 years agoRelated Solutions
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