Question 1185473
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.


<a href = "https://stattrek.com/probability-distributions/binomial.aspx" target = "_blank">https://stattrek.com/probability-distributions/binomial.aspx</a>