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

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

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 =
`(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(
`(X-\mu)/(\sigma)` <
`(3.1-3)/(0.05)` ) = P(Z < 2) = 0.97725

P(X
`\leq` 2.95 ounces) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(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.