z score homework

Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 31 liters, and standard deviation of 2.1 liters.

What is the probability that daily production is between 28.8 and 36.5 liters? Round to 4 decimal places.

Answer 1

x = mean production, mean 31, SD = 21.

want to find P(28.8 < x < 36.5)

use z = (x - mean)/SD = (x -31)/21 to transform x to standard normal distribution

so P(28.8 < x < 36.5) =

P( (28.8-31)/21 < z < (36.5-31)/21) =

P(-0.104 < z < 0.262) =

P(z < 0.262) - P(z < -0.104)

Lookup these probabilities from standard normal probability table