A random sample is to be selected from a population that has a proportion of successes p=.60. When n=400, what is the probability that a sample proportion falls between .59 and .62? Round your answer to three decimal places.

0.45134

Step-by-step explanation:

Given that :

p = 0.6

n = 400

Probability that sample. Proportion falls between 0.59 and 0.62

Using Normal approximation :

Mean (m) = n * p = 400 * 0.6 = 240

Standard deviation (s) = sqrt(pq/n)

q = 1 - p = 1 - 0.6 = 0.4

s = sqrt((0.6 * 0.4) / 400) = 0.0244948

P(0.59 < p < 0.62) :

(x - m) / s

P((0.59 - 0.6) / 0.0244948) < p < P((0.62 - 0.6) / 0.0244948)

P(Z < −0.408249) < p < P(Z < 0.8164998)

Using the Z probability calculator :

0.79289 - 0.34155 = 0.45134

so, converting firstly the mixed fractions to improper fractions and then "adding".