A certain car model has a mean gas mileage of 30 miles per gallon (mpg) with a standard deviation A pizza delivery company buys 57 of these cars. What is the probability that the average mileage of the fleet is greater than
30.3 mpg.
It would help if you proofread your questions before you posted them.
No standard deviation.
The standard deviation is 3
To calculate the probability that the average mileage of the fleet is greater than 30.3 mpg, we can use the concept of sampling distribution and the Central Limit Theorem.
The Central Limit Theorem states that if you have a large enough sample size, the sampling distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution. This means that as long as our sample size is large enough, we can assume that the distribution of the average mileage of the fleet is approximately normal.
In this case, the sample size is 57, which is considered large enough for the Central Limit Theorem to apply.
To calculate the probability, we need to standardize the average mileage value (30.3 mpg) using the formula for z-scores:
z = (x - μ) / (σ / sqrt(n))
where x is the sample mean (30.3 mpg), μ is the population mean (30 mpg), σ is the population standard deviation, and n is the sample size (57).
Since the standard deviation (σ) of the population is not given in the provided information, we cannot calculate the exact probability. However, if we assume a reasonable standard deviation value, we can provide an example calculation for illustration purposes.
Let's say we assume a standard deviation of 2.5 mpg. Now we can calculate the z-score:
z = (30.3 - 30) / (2.5 / sqrt(57))
Using a z-table or a statistical software, we can find the area under the standard normal distribution curve to the right of the calculated z-score. This area represents the probability that the average mileage of the fleet is greater than 30.3 mpg.
Please note that if you have the actual population standard deviation, you should use that value for a more accurate calculation.