Question

The following box model represents the population of those that find their work stressful: [101111101]. Given a sample of 200, what is the probability that more than 150 people find their job stressful? Assume a normal distribution.

Suppose we have a box model representing a die: [1,2,3,4,5,6] with a mean of μ=3.5 and a standard deviation of σ=1.708. If we do 88 rolls of the die what is the probability that the sum will be less than 300?

Answer 1

Answer:

1.

Given,

n = 200

p = 7/9

= 0.7778

mean = np

= 200*0.7778

= 155.56

standard deviation = sqrt(npq)

= sqrt(200*0.7778*(1-0.7778)

s = 5.88

Now P(X > 150) = P(z > (150 - 155.56)/5.88)

= P(z > -0.9456)

P(X > 150) = 0.8279 [since from z table]

P(X > 150) = 82.79%

2.

Mean = 3.5

s = 1.708

For 88 rolls of die , mean = 88*3.5

= 308

s = sqrt(n*s^2)

= sqrt(88*1.708^2)

s = 16.02

Now consider,

P(X < 300) = P(z < (300 - 308)/16.02)

= P(z < -0.5)

= 0.3085 [since from z table]

P(X < 300) = 0.3085