Question

A jar of peanut butter contains 454 g with a standard deviation of 10.2 g. Find the probability that a jar contains more than 466 g. Assume a normal distribution. Use a z-score rounded to 2 decimal places.

Answer 1

The probability that a jar contains more than 466 g is 0.119.

Explanation:

We are given that a jar of peanut butter contains a mean of 454 g with a standard deviation of 10.2 g.

Let X = Amount of peanut butter in a jar

The z-score probability distribution for the normal distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean = 454 g

`\sigma` = standard deviation = 10.2 g

So, X ~ Normal(
`\mu=454 , \sigma^(2) = 10.2^(2)`)

Now, the probability that a jar contains more than 466 g is given by = P(X > 466 g)

P(X > 466 g) = P(
`(X-\mu)/(\sigma)` >
`(466-454)/(10.2)` ) = P(Z > 1.18) = 1 - P(Z
`\leq` 1.18)

= 1 - 0.881 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.