SOLUTION: Assume that a procedure yields a binomial distribution with n=330 trials and the probability of success for one trial is p=0.19, Find the mean for this binomial distribution. (Rou

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Question 1185473: Assume that a procedure yields a binomial distribution with n=330 trials and the probability of success for one trial is p=0.19, Find the mean for this binomial distribution. (Round answer to one decimal place.)
μ=62.7
Find the standard deviation for this distribution.(Round answer to two decimal places.)
σ=7.13
Use the range rule of thumb to find the minimum usual value μ-2σ and the maximum usual value μ+2σ.
Enter answer as an interval using square-brackets only with whole numbers.
usual values=?
Find the usual values?

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
p = .19
q = 1 - 0 = .81
n = 330 = sample size

m = n * p = .19 * 330 = 62.7 = mean of the binomial distribution.
s = square root (n * p * q) = sqrt(330 * .19 * .81) = 7.12649999842 = standard deviation of the binomial distribution.

range rule of thumbs says:

68% of the scores are within 1 standard deviation of the mean.
95% of the scores are within 2 standard deviations of the mean.
99.7% of the scores are within 3 standard deviations of the mean.

you are looking at 95% of the scores being within 2 standard deviations of the mean.

the critical z-score for 95% of the scores being within 2 standard deviations of the mean would be plus of minus 1.96.

use the z-score formula to find the raw score.

z-score formula is:

z = (x - m) / s

z is the z-score
x is the raw score
m is the mean
s is the standard deviation

you get:

-1.96 = (x - 62.7) / 7.1265 for the low side.
1.96 = (x - 62.7) / 7.1265 for the high side.

solve for x in each to get:

x = -1.96 * 7.1265 + 62.7 = 48.73206 for the low side.
x = 1.96 * 7.1265 + 62.7 = 76.66794 for the high side.

the scores with 95% of the normal distribution between them would be 48.73206 to 76.66794.

if you round to the nearest integer, you get 49 to 77.

in interval notation that would be [49,77].

i don't know what is meant by usual values.
i think the above is your answer.
give it a whirl; see how you do.

here's a reference on binomial distribution.

https://stattrek.com/probability-distributions/binomial.aspx