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

Answer:

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

Step-by-step 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  =  $\frac{X-\mu}{\sigma}$  ~ 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( $\frac{X-\mu}{\sigma}$ > $\frac{466-454}{10.2}$ ) = 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.