Question

A) A particular fruit's weights are normally distributed, with a mean of 483 grams and a standard deviation of 21 grams.
If you pick one fruit at random, what is the probability that it will weigh between 479 grams and 485 grams?

B) A particular fruit's weights are normally distributed, with a mean of 478 grams and a standard deviation of 28 grams.
The heaviest 6% of fruits weigh more than how many grams? Give your answer to the nearest gram.

Answer 1

Solution :

Given that ,

A)

mean = = 483

standard deviation = = 21

P(479 < x < 485) = P[(479 - 483)/ 21) < (x - ) /  < (485 - 483) / 21) ]

= P(-0.19 < z < 0.10)

= P(z < 0.10) - P(z < -0.19)

= 0.5398 - 0.4247

= 0.1151

Probability = 0.1151

B)

mean = = 478

standard deviation = = 28

Using standard normal table ,

P(Z > z) = 6%

1 - P(Z < z) = 0.06

P(Z < z) = 1 - 0.06

P(Z < 1.56) = 0.94

z = 1.56

Using z-score formula,

x = z * +

x = 1.56 * 28 + 478 = 522

522 grams