Question

The amount of juice in a container is

normally distributed with a mean of

70 ounces and a standard deviation of

0.5 ounce. What is the probability that a

randomly selected container has more

than 70.5 ounces of juice?

Answer 1

0.15866

Explanation:

To solve for the above question, we use z score formula

z = (x - μ) / σ, where

x is the raw score

μ is the population mean

σ is the population standard deviation.

From the question,

x is the raw score = 70.5

μ is the population mean = 70

σ is the population standard deviation = 0.5

z = 70.5 - 70/ 0.5

z= 0.5/0.5

z = 1

We used the z score table to find the probability of z = 1

P( x = z) = P(x = 70.5)

= P( z = 1) = 0.84134

Hence, the probability that a randomly selected container has more than 70.5 ounces of juice is calculated as

P(x>70.5) = 1 - P(x =70.5)

P(x>70.5) = 1 - 0.84134 = 0.15866

Therefore, the probability that a randomly selected container has more than 70.5 ounces of juice is 0.15866.