Question

One out of four American adults has eaten Pizza for breakfast. If a sample of 20 adults is selected.

1. What type of probability distribution is this situation? ______________

2. None of them has eaten pizza for breakfast. ______________

3. What is the probability of less than 5 have eaten Pizza for breakfast? ___________

4. What is the probability have eaten between 5-10 Pizza for breakfast? ____________

5. What is the probability that more than 9 have eaten Pizza for breakfast? __________

6. Determine the mean of this sample? _____________

7. Determine the standard deviation of this sample _____________
8. How many adults do you expect to eat Pizza in this sample ____________

Answer 1

a) The number of people eating pizza for breakfast is modelled here as a binomial distribution given as:

b) The probability that none of them ate pizza for breakfast is computed here as:
P(X = 0) = (1 - 0.25)²⁰ = 0.0032

therefore 0.0032 is the required probability here.

c) The probability that less than 5 have eaten pizza for breakfast is computed here as:
P(X < 5)

This is computed in EXCEL as:
=binom.dist(4,20,0.25,TRUE)

The output here is 0.4148

Therefore 0.4148 is the required probability here.

d) The probability that number of people who have eaten pizza is between 5-10 is computed here as:
P( 5 <= X <= 10)
= P(X <= 10) - P(X <= 4)

This is computed in EXCEL here as:
=binom.dist(10,20,0.25,TRUE)-binom.dist(4,20,0.25,TRUE)

0.5812 is the output here.

Therefore 0.5812 is the required probability here.

e) The probability that more than 9 have eaten pizza is computed here as:
P(X > 9) = 1 - P( X <= 9)

This is computed in EXCEL as:
=1-binom.dist(9,20,0.25,TRUE)

0.0139 is the output here.

Therefore 0.0139 is the required probability here.

f) The mean of the sample is computed as:
Mean = np = 20*0.25 = 5

therefore 5 is the required mean here.

g) The standard deviation here is computed as: