Question

1. Whitney Gourmet Cat Food has determined the weight of their cat food can is normally distributed with a mean of 3 ounces and a standard deviation of 0.05 ounces. To meet legal and customer satisfaction goals each can must weigh between 2.95 and 3.1 ounces. a. If a single can is chosen, what is the probability it will weigh less between 2.95 and 3.1 ounces

Answer 1

Answer:

The probability it will weigh between 2.95 and 3.1 ounces is 0.8186.

Step-by-step explanation:

We are given that Whitney Gourmet Cat Food has determined the weight of their cat food can is normally distributed with a mean of 3 ounces and a standard deviation of 0.05 ounces.

Let X = weight of their cat food can

So, X ~ Normal($\mu=3,\sigma^{2} =0.05^{2}$)

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

                                Z =  $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 3 ounces

            $\sigma$ = standard deviation = 0.05 ounces

Now, the probability it will weigh between 2.95 and 3.1 ounces is given by = P(2.95 ounces < X < 3.1 ounces)

P(2.95 ounces < X < 3.1 ounces) = P(X < 3.1 ounces) - P(X $\leq$ 2.95 ounces)

   P(X < 3.1 ounces) = P( $\frac{X-\mu}{\sigma}$ < $\frac{3.1-3}{0.05}$ ) = P(Z < 2) = 0.97725

    P(X $\leq$ 2.95 ounces) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{2.95-3}{0.05}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)

                                                              = 1 - 0.84134 = 0.15866

The above probabilities is calculated by looking at the value of x = 2 and x = 1 in the z table which has an area of 0.97725 and 0.84134 respectively.

Therefore, P(2.95 ounces < X < 3.1 ounces) = 0.97725 - 0.15866 = 0.8186

Hence, the probability it will weigh between 2.95 and 3.1 ounces is 0.8186.