Probability that sample of 60 homes yields less than 2 or greater than 2.5 cars

Question

given a mean of 2.28 cars per house and standard deviation of 1.3

Answer 1

Find the z-scores according to z=(x-µ)/(σ/√n):

 

z₁=(2-2.28)/(1.3/√60)=-1.69

z₂=(2.5-2.28)/(1.3/√60)=1.31

 

Then use a table or the TI-84 to find

 

P(z≤-1.69)=.0476

 

P(z≥1.31)=.0951

 

Total probability is the sum,

 

P=.0476+.0951=.143

 

Answer 2

We want P(x<2) + P(x>2.5). We are given a mean and standard deviation. I think that what the question means to ask is the probability that the sample mean lies in the specified ranges. Otherwise the question doesn't make sense because you can't count between 2 and 2.5 cars. I assume that cars per household is an independent random variable from a normal distribution. If not, the Central Limit Theorem, shows that our answer will approximate the theoretical probability as long as the independence assumption holds.

 

The sample mean will be normally or nearly normally distributed with a mean of 2.28.

The standard deviation of the sample mean distribution will be the population standard deviation divided by the square root of the sample size.

 

μ=2.28

σ=1.3/sqrt(60)≈0.168

 

P(x<2) = Φ(z₁) 

z₁ = (x-μ)/σ = (2-2.28)/0.168 =-1.67

Φ(-1.67)=1-Φ(1.67)

Using a table, calculator, or computer we find that Φ(1.67) =0.9424 so 1-Φ(1.67) = 0.0476

 

 

P(x>2.5) = 1- P(x<2.5) = 1 -Φ(z₂)

z₂=(2.5-2.28)/0.168=1.31

Using a table, calculator, or computer we find that Φ(1.31) =0.9050 so 1-Φ(1.31) = 0.0950

 

So the probability that the sample mean is less than 2 or greater than 2.5 is 0.0476+0.0950=14.26%