Question

a particular fruit's weights are normally distributed, with a mean of 717 grams and a standard deviation of 18 grams. if you pick 12 fruits at random, then 16% of the time, their mean weight will be greater than how many grams? give your answer to the nearest gram.

Answer 1

If the mean is 7171 grams, the standard deviation is 18 grams and 12 fruits are selected at random. Then, the mean weight will be 734.901 grams greater.

Let X be the random variable of the fruit's weight.

We have, the mean and standard deviation as 717 grams and 18 grams respectively.

We want to find the specific gram value for which 16% of the fruits are heavier, i.e. the 84th percentile of X's normal distribution

Now, Solving for x:

P( X < x ) = Normal( 717, 18 ), Z = 0.84

So,

P( [X - 717] / 18 < [x - 717] / 18 ) = P( Z < [x - 717]/18 ) = 0.84, where Z is a standard normal random variable

Now,

Standard Normal CDF( [x - 717]/18 ) = 0.84

Then, from the standard normal table: [x - 717]/18 = Inverse Standard Normal CDF( 0.84 ) = 0.9945

Therefore,

[ x - 717 ] / 18 = 0.9945

Multiplying 18 on each side of the equation,

x - 717 = 0.9945 × 18

x - 717 = 17.901

Adding 717 on each side of the equation,

x - 717 + 717 = 17.901 + 717

x = 734.901

Learn more about standard deviation here:

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