The percent of births to mothers with less than a college education in all of the most populated cities in the U. S. has an average of 39.3 with a standard deviation of 4.1. What is the probability that in a random sample of six cities, the sample mean would be more than 40?

Probability that in a random sample of six cities, the sample mean would be more than 40 is 0.3372.

Step-by-step explanation:

We are given that the percent of births to mothers with less than a college education in all of the most populated cities in the U.S. has an average of 39.3 with a standard deviation of 4.1.

Assuming the data follows distribution. Also, a random sample of six cities is selected.

Firstly, Let = sample mean of six cities

The z score probability distribution for sample mean is given by;

          Z = ~ N(0,1)

where, = average percent of births = 39.3

             = standard deviation = 4.1

            n = sample of cities = 6

Probability that in a random sample of six cities, the sample mean would be more than 40 is given by = P( > 40)

  P( > 40) = P( > ) = P(Z > 0.42) = 1 - P(Z 0.42)

                                                  = 1 - 0.66276 = 0.3372

Therefore, probability that the sample mean would be more than 40 is 0.3372.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: