Question

the distribution of weights for widgets produced by a new machine is normal with a mean of 10 oz and a standard deviation of 2oz
a. what is the probability that one widget choosen at random from a days production will weigh between 9 and 11 ounces?
b. what is the probability that the average weight of a sample of 25 widgets chosen at random from a days production will weigh between 9 and 11 ounces?
what is the difference between the 2 problems?

Answer 1

solution:-

given that mean = 10 , standard deviation = 2

a. P(9 < x < 11) = P((9-10)/2) < z < (11-10)/2)

= P(-0.5 < z < 0.5)

= P(z < 0.5) - P(z < -0.5)

= P(z < 0.5) - (1-P(Z < 0.5))

= 0.6915 - (1-0.6915)

= 0.3830

b. here given that sample n = 25

=> P(9 < x < 11) = P((9-10)/(2/sqrt(25)) < z < (11-10)/(2/sqrt(25)))

= P(-2.5 < Z < 2.5)

= P(z < 2.5) - P(z < -2.5)

= P(z < 2.5) - (1-P(z < -2.5))

= 0.9938 - (1-0.9938)

= 0.9876

=> the difference between the 2 problems

from that below answer

part a = 0.3830

part b = 0.9876

by using the sample of 25 , the probability of part b is increased