Probability Distribution of Bedrooms: 1022386

Part (a) Probability distribution for the number of bedrooms

Part (b) Mean of this distribution.

Part (c) The standard deviation of this distribution

Question 7

Part (a) Probability that all six arrive within two days (P (x=6))

P(x=6) = _n C_x π^x ₍1- π) ^n-x

=₆C₆ (0.95)⁶ (1-0.95)^6-6

= 0.735 (Using binomial probability distribution table)

Part (b): Probability that exactly five arrive within two days (P (x=5))

P(x=5) = _n C_x π^x (1- π) ^n-x

=₆C₅ (0.95)⁵ (1-0.95)^6-5

= 0.232 (Using binomial probability distribution table)

Part (c): The mean number of letters that will arrive within two days

μ=nπ

=6*0.95

=5.7

Part (d) The variance that numbers of letters that will arrive within two days

σ²= nπ (1- π)

=6*0.95*(1-0.5)

=0.285

Standard deviation will be given by

σ=√ σ²2

=√0.285

=0.534

Question 8

μ = 221.10, σ = 47.11, x = 280

z = (x – μ)/σ = (280 – 221.10)/47.11 = 1.2503

P(x > $280) = P(z > 1.2503) = 0.1056

Number of homes selling for more than $280 = 105 * 0.1056 = 11.09  

Actual number of homes selling for more than $280 = 14

The normal distribution does not give a good approximation.