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

395, 17.212

(360.576,429.424)

Step-by-step explanation:

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

Here

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

Mean = np = 395

Variance = npq =

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

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation:

900 degrees

step-by-step explanation: