Question

(1 point) The distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.8 ounces and standard deviation 0.15 ounces. (a) What is the probability that the average weight of a bar in a Simple Random Sample (SRS) with five of these chocolate bars is between 7.64 and 7.9 ounces?

Answer 1

Answer:

Probability that the average weight of a bar in a Simple Random Sample (SRS) with five of these chocolate bars is between 7.64 and 7.9 ounces is 0.9235.

Step-by-step explanation:

We are given that the distribution of actual weights of 8-oz chocolate bars produced by a certain machine is normal with mean 7.8 ounces and standard deviation 0.15 ounces.

A simple random sample of five of these chocolate bars is taken.

Let $\bar X$ = sample average weight of a bar

The z-score probability distribution for sample mean is given by;

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

where, $\mu$ = average weight = 7.8 ounces

            $\sigma$ = standard deviation = 0.15 ounces

             n = sample of chocolate bars = 5

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, probability that the the average weight of a bar in a Simple Random Sample (SRS) with five of these chocolate bars is between 7.64 and 7.9 ounces is given by = P(7.64 ounces < $\bar X$ < 7.90 ounces)

P(7.64 ounces < $\bar X$ < 7.90 ounces) = P($\bar X$ < 7.90) - P($\bar X$ $\leq$ 7.64)

  P($\bar X$ < 7.90) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ <  $\frac{ 7.90-7.80}{{\frac{0.15}{\sqrt{5} } }} }$ ) = P(Z < 1.49) = 0.93189  {using z table}

  P($\bar X$ $\leq$ 7.64) = P( $\frac{ \bar X -\mu}{{\frac{\sigma}{\sqrt{n} } }} }$ $\leq$  $\frac{ 7.64-7.80}{{\frac{0.15}{\sqrt{5} } }} }$ ) = P(Z $\leq$ -2.39) = 1 - P(Z < 2.39)

                                                         = 1 - 0.99158 = 0.00842

The above probability is calculated by looking at the value of x = 2.39 and x = 1.49 in the z table which has an area of 0.99158 and 0.93189 respectively.

Therefore, P(7.64 ounces < $\bar X$ < 7.90 ounces) = 0.93189 - 0.00842 = 0.9235.