It is generally believed that electrical problems affect about 14% of new cars. An automobile mechanic conducts diagnostic tests on 128 new cars on the lot.
a. Describe the sampling distribution for sample proportion by naming the model and telling its mean and standard deviation. Justify your answer.
b. What is the probability that in this group over 18% of the new cars will be found to have electrical problems?
a. The sampling distribution for sample proportion can be modeled using the binomial distribution. The mean of the sampling distribution is the same as the population proportion, which is 14% or 0.14. The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:
Standard Error = sqrt((p * (1 - p)) / n)
Where p is the population proportion and n is the sample size.
In this case, p = 0.14 and n = 128. Plugging these values into the formula, we get:
Standard Error = sqrt((0.14 * (1 - 0.14)) / 128) ≈ 0.031
Therefore, the sampling distribution can be modeled as a binomial distribution with a mean of 0.14 and a standard deviation of 0.031.
b. To find the probability that over 18% of the new cars will be found to have electrical problems, we need to calculate the probability of getting a sample proportion greater than 0.18.
Using the sampling distribution, we can calculate the z-score for a sample proportion of 0.18:
z = (x - μ) / σ
Where x is the sample proportion, μ is the mean of the sampling distribution, and σ is the standard deviation of the sampling distribution.
Plugging in the values, we get:
z = (0.18 - 0.14) / 0.031 ≈ 1.29
Using a standard normal distribution table or calculator, we can find the probability of getting a z-score greater than 1.29.
The probability that in this group over 18% of the new cars will be found to have electrical problems is approximately 0.0985 or 9.85%.
a. To describe the sampling distribution for sample proportion, we use a binomial distribution model because we're interested in the number of new cars with electrical problems out of the 128 tested.
The mean (μ) of the binomial distribution is given by n * p, where n is the sample size (128 in this case) and p is the probability of electrical problems in a new car (14%).
μ = 128 * 0.14 = 17.92
The standard deviation (σ) of a binomial distribution is given by √(n * p * (1 - p)).
σ = √(128 * 0.14 * (1 - 0.14)) ≈ 3.26
Therefore, the sampling distribution for sample proportion follows a binomial distribution with a mean of 17.92 and a standard deviation of 3.26.
b. To find the probability that over 18% of the new cars will be found to have electrical problems, we need to calculate the probability of selecting more than a certain number of cars with electrical problems.
We can use the normal approximation to the binomial distribution since the sample size is large (n = 128) and np > 10 and n(1-p) > 10.
First, we need to find the z-score for 18%. The formula for the z-score is:
z = (x - μ) / σ,
where x is the desired proportion (18%), μ is the mean of the distribution (17.92), and σ is the standard deviation of the distribution (3.26).
z = (0.18 - 0.14) / 0.0326 ≈ 1.23
Next, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator. Let's assume we're interested in finding the probability of selecting more than 18% (i.e., right-tail probability).
P(Z > 1.23) ≈ 1 - P(Z < 1.23)
Using a standard normal distribution table or calculator, we can find that P(Z < 1.23) is approximately 0.8907.
Therefore, P(Z > 1.23) ≈ 1 - 0.8907 ≈ 0.1093.
Therefore, the probability that in this group over 18% of the new cars will be found to have electrical problems is approximately 0.1093 or 10.93%.
This is a binominal distribution -- either the car has problems or it doesnt.
With n=128, the expected mean number with problems is .14*128=17.92 The standard deviation is sqrt(p*q*n) = sqrt(.14*.86*128) = 3.9
.18*128=23.04, which is 5.12 above the mean or 5.12/3.9=1.31 standard deviations away from the mean.
Take it from here....