Question

1)A particular fruit's weights are normally distributed, with a mean of 543 grams and a standard deviation of 27 grams. If you pick 12 fruit at random,

what is the probability that their mean weight will be between 531 grams and 537 grams?

2)

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 157.7-cm and a standard deviation of 1.9-cm. For shipment, 22 steel rods are bundled together.

Find P₆₃, which is the average length separating the smallest 63% bundles from the largest 37% bundles.

P₆₃ =  cm Round to 2 decimal places.

Answer 1

1. solution:-

given that mean = 543 , standard deviation = 27 , n = 12

=> P(531 < x < 537)

=> P((531-543)/(27/sqrt(12)) < z < (537-543)/(27/sqrt(12)))

=> P(-1.54 < z < -0.77)

=> P(z < -0.77) - P(z < -1.54)

=> (1-0.7794) - (1-0.9382)

=> 0.2206 - 0.0618

=> 0.1588

2. solution:-

given that mean = 157.7 , standard deviation = 1.9 , n = 22

the z value that corresponds to the probability of 0.63 from standard normal table is

z = 0.332

formula

=> z = (x-mean)/(sd/sqrt(n))

=> x = z*(sd/sqrt(n)) + mean

=> x = 0.332*(1.9/sqrt(22))+157.7

=> x = 157.83