Question

Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. Refer to Exhibit 6-5. What percentage of items will weigh between 6.4 and 8.9 ounces

Answer 1

Answer:

46.18% of the items will weigh between 6.4 and 8.9 ounces.

Step-by-step explanation:

We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

Let X =  weight of items produced by a machine

The z-score probability distribution for is given by;

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

where, $\mu$ = mean weight = 8 ounces

            $\sigma$ = standard deviation = 2 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X $\leq$ 6.4 ounces)

   P(X < 8.9) = P( $\frac{ X -\mu}{\sigma}$ < $\frac{ 8.9-8}{2}$ ) = P(Z < 0.45) = 0.67364  {using z table}

   P(X $\leq$ 70) = P( $\frac{ X -\mu}{\sigma}$ $\leq$ $\frac{ 6.4-8}{2}$ ) = P(Z $\leq$ -0.80) = 1 - P(Z < 0.80)

                                                 = 1 - 0.78814 = 0.21186

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.45 and x = 0.80 in the z table which has an area of 0.67364 and 0.78814 respectively.

Therefore, P(6.4 < X < 8.9) = 0.67364 - 0.21186 = 0.4618 or 46.18%

Hence, 46.18% of the items will weigh between 6.4 and 8.9 ounces.