Having trouble knowing where to start with this one:
A maker of tools claims the average amount of time it takes to assemble a table saw is 50 mins with a standard deviation of 40 mins. What is the probability that the sample mean will be less than 44 mins if a random sample of 64 buyers of the saw is taken?
Thanks for any help!
To solve this problem, we will use the concept of sampling distribution and the z-score.
1. First, let's define the variables:
- Population mean (μ) = 50 minutes
- Population standard deviation (σ) = 40 minutes
- Sample size (n) = 64
- Sample mean (x̄) = 44 minutes
2. Calculate the standard error of the mean (SE):
SE = σ / √n
SE = 40 / √64
SE = 40 / 8
SE = 5 minutes
3. Calculate the z-score:
The z-score measures the number of standard deviations a given value (in this case, the sample mean) is from the population mean.
z = (x̄ - μ) / SE
z = (44 - 50) / 5
z = -6 / 5
z = -1.2
4. Determine the probability:
We want to find the probability that the sample mean will be less than 44 minutes. This is equivalent to finding the area under the normal distribution curve to the left of the z-score.
To find this probability, we need to standardize the z-score and use a standard normal distribution table or a calculator.
Using the z-table or a calculator, we can determine that the probability corresponding to a z-score of -1.2 is approximately 0.1151. This means there is a 11.51% chance that the sample mean will be less than 44 minutes.
Therefore, the probability that the sample mean will be less than 44 minutes if a random sample of 64 buyers of the saw is taken is approximately 11.51%.