The mean annual salary for intermediate level executives is about $74000 per year with a standard deviation of $2500. a random sample of 50 intermediate level executives is selected. what is the probability that the mean annual salary of the sample is between $71000 and $73500? a 0.079b. 0.500c. 0.487d. 0.306

D. 0.306

Step-by-step explanation:

Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = annual salary for intermediate level executives

u = mean annual salary

s = standard deviation

From the information given,

u = $74000

s = $2500

We want to find the probability that the mean annual salary of the sample is between $71000 and $73500. It is expressed as

P(71000 lesser than or equal to x lesser than or equal to 73500)

For x = 71000,

z = (71000 - 74000)/2500 = - 1.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.1151

For x = 73500,

z = (73500 - 74000)/2500 = - 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.4207

P(71000 lesser than or equal to x lesser than or equal to 73500) is

0.4207 - 0.1151 = 0.306

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)