Concept explainers
(a)
To compute: Mean and standard deviation of the weight of a box of candy.
(a)
Answer to Problem 34PT3
Mean
Standard deviation
Explanation of Solution
Given information:
Number of chocolate truffles in a gift box
Number of handmade caramel nougats
Mean of a truffle
Standard deviation of a truffle
Mean of nougat
Standard deviation of nougat
Mean of an empty box
Standard deviation of an empty box
Formula used:
If independent
Calculation:
Mean of 8 truffles
Variance of 8 truffles
Mean of 2 nougats
Variance of 2 nougats
As the truffle, nougat and empty box are independent, so
The mean of a candy box is
Variance of a candy box
Standard deviation of a candy box =
(b)
To compute: Probability that a candy box selected will weigh more than 30 ounces.
(b)
Answer to Problem 34PT3
Probability that a candy box selected will weigh more than 30 ounces is 0.06772.
Explanation of Solution
Given information:
The weights of truffles, nougats and empty box are normally distributed.
Formula used:
Z score:
Calculation:
Let X the weight of a candy box.
The distribution of will be approximately normal due to the additive property of normal distributions.
So,
From z tables,
Using linear interpolation
Hence,
(c)
To compute: Probability that at least one of the 5 boxes selected will weigh more than 30 ounces.
(c)
Answer to Problem 34PT3
Probability that at least one of the 5 boxes selected will weigh more than 30 ounces is 0.2957.
Explanation of Solution
Formula used:
Calculation:
As each gift box has same probability of weighing more than 30 ounces and are independent, so it can be modelled by a binomial distribution.
Let Y be the number of gift boxes weighing more than 30 ounces.
So,
Hence, the required probability is
(d)
To compute: Probability that the mean weight of five boxes is more than 30 ounces.
(d)
Answer to Problem 34PT3
Probability that the mean weight of five boxes is more than 30 ounces is 0.000423.
Explanation of Solution
Concept used:
Central limit theorem
Calculation:
From z tables,
Using linear interpolation
Hence, the required probability is
Chapter 10 Solutions
The Practice of Statistics for AP - 4th Edition
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