The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is mound-shaped and symmetric with a mean of 24.6 mpg and a standard deviation of 11.2 mpg. If 30 such cars are tested, what is the probability the average mpg achieved by these 30 cars will be greater than 27
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the probability that the average mpg achieved by these 30 cars will be greater than 27, we need to use the concept of sampling distribution and the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size (n ≥ 30), the sampling distribution of the sample mean is approximately normally distributed, regardless of the shape of the population distribution. In this case, we can assume that the average mpg of the 30 cars follows a normal distribution.
To calculate the probability, we need to standardize the average mpg (x̄) using the formula for the z-score:
z = (x̄ - μ) / (σ / sqrt(n))
Where:
x̄ is the sample mean (27 in this case),
μ is the population mean (24.6 mpg),
σ is the population standard deviation (11.2 mpg),
n is the sample size (30).
Plug in the given values into the formula:
z = (27 - 24.6) / (11.2 / sqrt(30))
Calculating this gives us the z-score. Next, we need to determine the corresponding probability from the standard normal distribution table or by using a statistical software or calculator.
For example, if we find that the z-score is 1.5, we can look up the corresponding probability (p) in a standard normal distribution table. Assuming a two-tailed test, we find that the probability of obtaining a z-score of 1.5 or greater is 0.0668.
So, the probability that the average mpg achieved by these 30 cars will be greater than 27 is approximately 0.0668 or 6.68%.