A particular fruit's weights are normally distributed, with a mean of 335 grams and a standard deviation of 38 grams. The heaviest 2% of fruits weigh more than how many grams? Give your answer to the nearest gram.
Solution :
mean = = 335
standard deviation = = 38
Using standard normal table,
P(Z > z) = 2%
1 - P(Z < z) = 0.02
P(Z < z) = 1 - 0.02 = 0.98
P(Z < 2.054) = 0.98
z = 2.054
Using z-score formula,
x = z * +
x = 2.054 * 38 + 335 = 413
Answer = 413 grams
SOLUTION :
Let a grams be the weight over which 2% fruits are there.
P( x > a) = P( z > (a - m)/s)
=> 0.02 = P( z > (a - m)/s)
From ND table :
z > 2.054
The cutoff point is z = 2.054
So,
a = z * s + m = 2.054 * 38 + 335 = 413 grams
So, the heaviest 2% fruits weigh more than 413 grams (ANSWER)
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