Question

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=1580,=1/4

Answer 1

395, 17.212

(360.576,429.424)

Explanation:

Given that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p.

Here
`n=1580and p =1/4 =0.25\\q = 0.75\\np = 395 and\\nq = 1185`

Since np and nq are greater than 5 by rule of thumb we can approximate binomial to normal.

Mean = np = 395

Variance = npq =
`395*0.75=296.25`

Std dev = 17.212

Thus X no of successes is N(395, 17.212)

THe the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma would be

`395-2(17.212), 395+2(17.212)\\= 360.576,429.424`