An automatic machine in a manufacturing process is operating properly if the length of an important subcomponent is normally distributed with a mean μ = 80 cm and a standard deviation σ = 2 cm.
a) Find the probability that the length of one randomly selected unit is less than 79 cm.
b) Find the probability that the average length of 10 randomly selected units is less than 79 cm.
c) Find the probability that the average length of 30 randomly selected units is less than 79 cm.
a) Z = (score-mean)/SD
b, c) 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 proportions/probabilities related to the Z scores.
To answer these questions, we need to use the properties of the normal distribution and z-scores. A z-score measures the number of standard deviations a data point is away from the mean.
a) Find the probability that the length of one randomly selected unit is less than 79 cm:
We can calculate the z-score using the formula: z = (x - μ) / σ, where x is the value we are interested in, μ is the mean, and σ is the standard deviation.
In this case:
x = 79 cm
μ = 80 cm
σ = 2 cm.
Plugging in these values, we get:
z = (79 - 80) / 2 = -0.5
Now, we need to find the probability of a z-score less than -0.5. We can consult the standard normal distribution table or use a calculator to find this probability.
Using a standard normal distribution table, we can find that the probability of a z-score less than -0.5 is approximately 0.3085. Therefore, the probability that the length of one randomly selected unit is less than 79 cm is approximately 0.3085.
b) Find the probability that the average length of 10 randomly selected units is less than 79 cm:
When dealing with the average of multiple random variables, the central limit theorem tells us that the distribution of the sample mean approaches a normal distribution as the sample size increases, regardless of the shape of the original distribution.
In this case, the distribution of the sample mean will have the same mean as the original distribution (μ = 80 cm) but a smaller standard deviation, which is the standard deviation of the original distribution divided by the square root of the sample size (σ / √n).
We are interested in finding the probability that the average length of 10 randomly selected units is less than 79 cm. So now we have:
x = 79 cm
μ = 80 cm
σ = 2 cm
n = 10
Calculating the standard deviation of the sample mean:
σ / √n = 2 cm / √10 ≈ 0.632 cm.
Now we can use the z-score formula again:
z = (x - μ) / (σ / √n)
= (79 - 80) / 0.632
= -1.5811
Using the standard normal distribution table or a calculator, we can find that the probability of a z-score less than -1.5811 is approximately 0.0571. Therefore, the probability that the average length of 10 randomly selected units is less than 79 cm is approximately 0.0571.
c) Find the probability that the average length of 30 randomly selected units is less than 79 cm:
Following the same steps as part b, we can calculate the standard deviation of the sample mean:
σ / √n = 2 cm / √30 ≈ 0.365 cm.
Calculating the z-score:
z = (79 - 80) / 0.365
= -2.7397
Using the standard normal distribution table or a calculator, we can find that the probability of a z-score less than -2.7397 is approximately 0.0032. Therefore, the probability that the average length of 30 randomly selected units is less than 79 cm is approximately 0.0032.