A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heaviest 5% of fruits weigh more than how many grams?

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 = weights of the fruits

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.

f(x) is the value of the function. b is the value of the function when x equals zero or the y-coordinate of the point where the line crosses the y-axis in the coordinate plane.

step-by-step explanation:

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