if a sample size 40 is drawn from the population that has mean 276 and a variance of 81, what is the probability that the mean of the sample will be less than 273?

The probability is 0.017429

Step-by-step explanation:

We start by calculating the z-score

Mathematically;

z-score = (x-mean)/SD/√n

In this case;

x = 273

mean = 276

SD = √(variance) = √81 = 9

n = 40

Substituting these values into the z-score formula, we have that;

z = (273-276)/9/√40

z = -2.11

So we proceed to get the probability

the probability in this case is;

P( z < -2.11)

We can use the normal distribution table for this

Our answer will be

P( z < -2.11) = 0.017429

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation: