Probability and Statistics Carpooling Question

Question

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 76% of vehicles on the roads are occupied by just the driver. (Round your answers to four decimal places.)
(a) If you choose 10 vehicles at random, what is the probability that more than half (that is, 6 or more) carry just one person?



(b) If you choose 98 vehicles at random, what is the probability that more than half (that is, 50 or more) carry just one person? (Use the normal approximation.)

Answer 1

The probability of a single rider is p = .76 and q = .24 the probability there are more than one rider in the car.

 

(a) Given n = 10 vehicles are checked, and that this is represented by a binomial distribution

 

Let P[k] = P[k cars have only a single rider] = ₁₀C_k * p^k *q^10-k

≈

P[More than half have carry a single rider] = P)6) + P(7) + P(8) + P(9) + P(10)

 

P(6) = ₁₀C₆ * p⁶ *q⁴ = 210 * (,76)⁶ * (.24)⁴ ≈ 0.134

P(7) = ₁₀C₇ * p⁷ *q³ = 120* (,76)⁷ * (.24)³ ≈ 0.243

P(8) = ₁₀C₈ * p⁸ *q²= 45 * (,76)8* (.24)² ≈ 0.288

P(9) = ₁₀C₉* p⁹ *q¹ = 10 * (,76)⁹ * (.24)¹ ≈ 0.203

P(10) = ₁₀C₁₀ * p¹⁰ *q⁰ = 1 * (,76)¹⁰ * (.24)⁰ ≈ 0.064

 

Adding these together, P[More than half have carry a single rider] ≈ 0.932

 

(a) Let n = 98

 

Is a normal approximation appropriate for a binomial distribution?

It is a good representation when np and nq greater than 10.

np = 98*.76 = 74,48 and nq = 98*.24 = 23.52

So this is a good approximation for the binomial distribution.

 

What is needed is the mean and standard deviation of this binomial distribution.

 

Mean: μ = np = 74.48

Standard deviation: σ = √(npq) ≈ 4.228

 

We want P[more than half the cars carry just one person]

Let X = 50, which is the value more than half.

 

To use the normal approximation a Z value must be determined.

The Z value is determined by Z = (X - μ)/σ, where the probability

that values less than Z can be found from a Standard Normal

Distribution Table.

In this case that probability is

P[less than half the cars carry just one person]

 

The Z value is Z = (50 - 74.48)/4.228 = -5.799

If Z < -3.89, then the probability of the event is 0 which is true here.

P[less than half the cars carry just one person] = 0

=> P[more than half the cars carry just one person] = 1