Question

What is the probability of selecting 50 cows with a sample mean of more than 1650 pounds? (show work)

X =

Z =

Probability = (Need to use Standard Normal Table)

Answer 1

x= 1650

z= -1.178511302

p= .119

Step-by-step explanation:

We have a normal distribution with mean 1600 and standard deviation 300

(if 300 is the variance then the z value is like -20 and the probability is 0)

To solve for the mean standard devation convert the standard deviation to a variance, divide by the number of terms, and then convert that to a standard deviation:

`\sqrt{((300^2)/(50))}= 42.42640687`

The average sample mean is just 1600

Solve for the Z value

`(1650-1600)/(42.42640687)=1.178511302`

because we're trying to solve for the probability of the cows being greater than a number we swap the sign

therefore the z value is

-1.178511302

Go to a z table, go to the value 1.18, and take the compliment

1-.8810=.119