Question

The number of marshmallows in a 24 oz. box of Lucky Charms Cereal is normally distributed with a mean of 100 marshmallows and a standard deviation of 15 marshmallows.

A) What is the probability that a box of cereal will have no more than 110 marshmallows in it? a. 0.75 b. 0.68 c. 0.83 d. 0

B) What is the probability that a box of cereal will have between 80 to 120 marshmallows? a. 0.32 b. 0.67 c. 0.82 d. 0.5

C) How many marshmallows must be in a box so that the box is in the top 3% with respect to marshmallow content? a. >106 b. >112 c. >128 d. >142

D) The manager wants 0.9 probability that a box of cereal has 60 or more marshmallows. The process remains normally distributed and the standard deviation of process remains the same (i.e., 15 marshmallows). What should be the mean of the number of marshmallows in a box of cereal? a. 85 b. 79 c. 105 d. 105

In this process the speed at which a box of cereal is filled and sealed is uniformly distributed and takes between 30 to 45 seconds.

A) What is the probability that a box can be filled and sealed in less than 42 seconds? a. 0.08 b. 0.6 c. 0.5 d. 0.25

B) What is the probability that the time it takes to fill and seal a box is between 40 to 50 seconds? a. 0.33 b. 0.50 c. 0.67 d. 0.25

Answer 1

This is a normal distribution question with

Mean() = 100 Standard Deviation (0) = 15 Since we know that score =

a) x = 110

P(x < 110.0)=?

The z-score at x = 110.0 is,

z = \frac{110.0-100.0}{15.0}

z = 0.6667

This implies that

P(x < 110.0) = P(z < 0.6667) = \textbf{0.7475181106016127}=0.75

b) x1 = 80

x2 = 120

P(80.0 < x < 120.0)=?

The 2 – score at r = 80.0is. 80.0 – 100.0 2 = 15.0 z1 = -1.3333 The 2 – score at r = 80.0is, 120.0 - 100.0 15.0 22 = 1.3333

This implies that

P(80.0 < x < 120.0) = P(-1.3333 < z < 1.3333) = P(Z < 1.3333) - P(Z < -1.3333)

P(80.0 < x < 120.0) = 0.9087833131501531 - 0.09121668684984685

P(80.0 < x < 120.0) = \textbf{0.8176}=0.82

c) Given in the question

top 3% means bottom 97%

P(X < x) = 0.97

This implies that

P(Z < 1.8807936081512509) = 0.97

With the help of formula for z, we can say that

Τ = μ + εσ + 1 = 100.0 + (1.8807936081512509) 15.0 1 = 128.2119

d) Given in the question

means data is separated from bottom 90%

P(X < x) = 0.1

This implies that

P(Z < -1.2815515655446004) = 0.1

With the help of formula for z, we can say that

= μ+ εσ 60 = μ + (-1.28)15.0 μ = 60 + 19.2 μ = 79.2

PS: you have to refer z score table to find t