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

the correct answer is d

hope this

found it xd it ok yt