We are given that the weight of potato chip bags filled by a machine at a packaging plant is normally distributed with a mean of 15.0 ounces and a standard deviation of 0.1 ounces.

Let X = weight of potato chip bags filled by a machine

So, X ~ N( $\mu=15.0,\sigma^{2} =0.1^{2}$ )

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

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

where, $\mu$ = mean weight = 15.0 ounces

$\sigma$ = standard deviation = 0.1 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 bags that weigh more than 14.8 ounces is given by = P(X > 14.8 ounces)

P(X > 14.8 ounces) = P( $\frac{ X -\mu}{\sigma}$ > $\frac{14.8-15.0}{0.1}$ ) = P(Z > -2) = P(Z < 2)

= 0.9773 or 97.73%

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 = 2 in the z table which has an area of 0.9773.

Hence, 97.73% of bags weigh more than 14.8 ounces.

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4 years ago