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A child who is trick-or-treating chooses a lollipop at random from a bowl of lollipops with mean weight 1.5 oz and standard deviation 0.3 oz. The child then chooses two chocolate bars at random from a bowl of bars with mean weight 2.0 oz and standard deviation 0.4 oz. The three choices are independent. Find the expected combined weight of the child's lollipop and two chocolate bars.

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Answer:

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

Explanation:

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Addition of normal variables:

The mean is the sum of the means.

Find the expected combined weight of the child's lollipop and two chocolate bars.

Child lollipop: Mean 1.5 oz

Chocolate bars: Mean 2 oz

One lollipop and 2 bars combined

The mean will be 1.5 + 2*2 = 5.5 oz

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

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