Question

Problem 1

It is known that the weights of M&M milk chocolate candy bags are normally distributed with the average weight of 11.04 ounces and the standard deviation of 0.82 ounces. One bag is randomly selected. Answer the following

Question 1 What is the probability that the weight of this bag is 10.65 ounces?

a) 0.3156

b) 0

c) 0.6844

d) none of the above

Question 2 What is the probability of the weight of this bag exceeding 11.93 ounces?

a) 0.1379

b) 0.4013

c) 0.8621

d) none of the above

Question 3 Find the probability that the weight of this bag is between 10.45 and 12.17 ounces

a) 0.7389

b) 0.4325

c) 0.6804

d) none of the above

Answer 1

Solution :

Given that ,

mean = = 11.04

standard deviation = = 0.82

Question 1

P(x < 10.65) = P[(x - ) / < (10.65 - 11.04) / 0.82]

= P(z < -0.48)

= 0.3156

probability = 0.3156

option a) is correct

Question 2

P(x > 11.93) = 1 - P(x < 11.93)

= 1 - P[(x - ) / < (11.93 - 11.04) / 0.82]

= 1 - P(z < 1.09)

= 1 - 0.8621

= 0.1379

Probability = 0.1379

option a) is correct

Question 3

P(10.45 < x < 12.17) = P[(10.45 - 11.04)/ 0.82) < (x - ) /  < (12.17 - 11.04) / 0.82) ]

= P(-0.72 < z < 1.38)

= P(z < 1.38) - P(z < -0.72)

= 0.9162 - 0.2358

= 0.6804

Probability = 0.6804

option c) is correct