At university A, 46% of the undergraduate students are female. At university B, 52% of the undergraduate students are female. Suppose a random sample of 100 students from both schools was selected. Assume the sampling distribution of p hat Subscript Upper A Baseline minus p hat Subscript Upper B is approximately Normal. What is the probability that the proportion of females selected from university A is greater than the proportion of females selected from university B?

If needed, use the z-table to answer the question.

Report your answer to 3 decimal places.

.198

Step-by-step explanation:

got it right on edg

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5