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Submit Question Question 8 A particular fruit's weights are normally distributed, with a mean of 618 grams and a standard deviation of 30 grams. If you pick one fruit at random, what is the probability that it will weigh between 519 grams and 615 grams Question Help Message instructor 2 Calculator Submit Question

Answer 1

Solution :

Given that,

mean = $\mu$ = 618

standard deviation = $\sigma$ = 30

P (519< x < 615 )

P (519 - 618 / 30) < ( x -  $\mu$/ $\sigma$ ) < ( 615 - 618 / 30 )

P ( -99 / 30 < z < - 3 / 30 )

P (-3.3 < z < - 0.1)

P ( z < - 0.1 ) - P ( z < -3.3)

Using z table

= 0.4602 - 0.0005

= 0.4597

Probability = 0.4597