Question

Please help me with these statistics questions, thank you

Answer 1

1. The table shows the probability of selling x houses in one day. So

P(X = 4) = 0.0551

2. Consult the table.

P(X = 0) = 0.2019

3. Add up the probabilities of selling at least 5 houses:

P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + …

By nature of the Poisson distribution, the probability of selling more than 9 houses is so close to 0 it may as well be. So we stop where the table does.

P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

= 0.0176 + 0.0047 + 0.0011 + 0.0002 + 0.0000

= 0.0236

4. As alluded in the previous answer,

P(X ≥ 10) ≈ 0

5. This requires some knowledge about the Poisson distribution. If an average of λ houses/day is sold, then the probability of selling x houses on a given day is

`P(X = x) = (\lambda^x e^(-\lambda))/(x!)`

If an average of 1.6 houses are sold each day, then in two days the average sold is

(1.6 houses) / (2 days) = 0.8 houses/day

and this follows another Poisson distribution with mean 0.8. Then the probability of selling 4 houses in two days is

`(0.8^4 e^(-0.8))/(4!) \approx \mathbf{0.0077}`