In: Advanced Math
The weight of apples is normally distributed. Large apples have a mean of 15 oz and medium apples have a mean of 10 oz. The standard deviation of the weight of both large and medium apples is 2 oz.
You select a large apple and a medium apple at random. Let L be the weight of the large apple and M be the weight of a medium apple.
Let X be the total weight of the two apples (X = L + M). The distribution of X is also normal.
1. Use the rules for means and variances to find the mean and standard deviation of X.
2. What is the probability that the total weight X is between 22 and 28 oz?
Let L be the weight of the large apples and M be the weight of medium apples.
Mean of large apples and medium apples are given as 15oz and 10oz respectively.
Thus and .
It is also given that X is the total weight (L+M)
But the mean of total weight of a large apple and medium apple is given as:
substituting the values:
.
Thus mean of total weight of the two apples is 25.
Now, to find standard deviation; remember that variance of sum of two random variables is the sum of variance of those random variables.
i.e. as it is given as both large and medium apples have standard deviation 2, they will have a variance 4. And thus the variance of total weight of the two apples will be:
.
Thus; standard deviation of the total weight of two apples will be;
.
Thus mean of X is 25 and variance of X is
.
2.
Now, consider the given question. The objective is to find the probability that total weight will be between 22 and 28.
So, let us find the z-values of 22 and 28 in the given distribution X at first:
z-value of 22 is:
i.e. .
Similarly z-value for 28 is:
.
Now; to find the probability that total weight is between 22 and 28, we just need to find the probability of z to be between -1.0607 and 1.0607 in a standard normal distribution.
i.e.
Using a normal distribution calculator .
(you can also use an appropriate normal distribution table)
Therefore the probability that total weight X is between 22 and 28 is: 0.7112
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