Question

Binomial distribution. Tables must be completed in excel only.

1. A report to Health Canada indicated that an intensive education program was used in an attempt to lower the rate of lung cancer development among females. After running the program, a long-term follow-up study of 505 females was conducted, which revealed the current rate of lung cancer development is 7.5 %.

1. Find the probability that 10 or less females have lung cancer development.
2. Assuming the program has effect, find the mean and standard deviation for the number of lung cancer cases in groups of 505 females.      
3. Considering that unusual values are outside the interval μ-2σ, μ+2σ, is 20 females out of 505 females with lung cancer development unusually low or not? Show calculations.   

Answer 1

Let x be the number of females have lung cancer development.

x follows binomial distribution with n = 505 and p = 0.075 , q = 1- p = 0.925

a) We are asked to find P( x ≤ 10 )

We can use Excel function =BINOM.DIST( x , n, p, cumulative ) to find the binomial probability.

We plug the values of x,n, p in the given order and for cumulative we plug "TRUE" because we have to find the sum of the probabilities from 0 to x

So P( x ≤ 10 ) = 3.5*10-8   ~ 0

Probability that 10 or less females have lung cancer development is approximately 0

b) Mean (µ) = n*p = 505*0.075 = 37.875

Standard deviation ( σ ) = =   = 5.9190

c)

μ -2σ = 37.875 - (2*5.9190) = 26.04

μ+2σ = 37.875 + (2*5.9190) = 49.71

Since the number 20 is less than 26.04, therefore 20 females out of 505 females with lung cancer development unusually low.