in the cafeteria ,100 mike cartons were put out for breakfast.At the end of breakfast,27 remainded,what is the ratio of the numb
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Answer:
The heaviest 5% of fruits weigh more than 747.81 grams.
Step-by-step explanation:
We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.
Let X = <u><em>weights of the fruits</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean weight = 733 grams
= standard deviation = 9 grams
Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;
P(X > x) = 0.05 {where x is the required weight}
P( >
) = 0.05
P(Z > ) = 0.05
In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;
x = 747.81 grams
Hence, the heaviest 5% of fruits weigh more than 747.81 grams.