The average age of a vehicle registered in the U.S. is 8 years, or 96 months. If a random sample of 36 vehicles is selected, find the probability that the mean of their age is between 98 and 100 months. Assume the standard deviation for the population is 15. (sample mean)
calculate the z scores:
z= (x-xbar)/(σ/sqrt(n))
where xbar=sample mean, σ=standard deviation of population, n=sample size, x=raw score (i.e. 98, 100)
now refer to the z chart:
http://www.thebeststatistics.info/sta/table_zvals.html
since you want the probability that the mean is between 98 and 100 months, I'd use the p-values on the right column (i.e. z-score to infinity) and take the difference of the p-values to get the area between 98 and 100.
Well, if we assume that the age of vehicles follows a normal distribution, we can use the Central Limit Theorem to approximate the distribution of the sample mean. The mean of the sample mean will still be 96 months, but the standard deviation of the sample mean will be 15 divided by the square root of the sample size (which is 36 in this case, since we have selected a random sample of 36 vehicles).
The standard deviation of the sample mean is 15 / √36 = 15 / 6 = 2.5 months.
To find the probability that the mean of their age is between 98 and 100 months, we need to calculate the z-scores for both values using the formula:
z = (x - μ) / σ,
where x is the value we're interested in (98 or 100), μ is the population mean (96), and σ is the standard deviation of the sample mean (2.5).
For 98 months:
z1 = (98 - 96) / 2.5 = 0.8
For 100 months:
z2 = (100 - 96) / 2.5 = 1.6
Now, we need to find the probability that the sample mean falls between the z-scores of 0.8 and 1.6, which is equivalent to finding the area under the standard normal curve between these z-scores.
We can use a standard normal distribution table or a calculator to find these probabilities. Using a standard normal distribution table, the probability corresponding to a z-score of 0.8 is 0.7881, and the probability corresponding to a z-score of 1.6 is 0.9452.
So, the probability that the mean of the vehicle ages is between 98 and 100 months is the difference between these two probabilities:
P(98 ≤ x ≤ 100) = 0.9452 - 0.7881 ≈ 0.1571.
Therefore, the probability is approximately 0.1571. Keep in mind that this is an approximation based on the assumptions made. Enjoy your data analysis with a touch of statistics humor!
To find the probability that the mean age of a random sample of 36 vehicles is between 98 and 100 months, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a sufficiently large sample size, the distribution of the sample means will be approximately normally distributed, regardless of the shape of the population distribution.
In this case, the sample size is 36, which is considered sufficiently large.
Step 1: Calculate the standard error of the mean.
The standard error of the mean, also known as the standard deviation of the sample mean, can be calculated using the formula:
Standard Error = Standard Deviation / √(sample size)
Given that the standard deviation for the population is 15 and the sample size is 36, we can plug in the values:
Standard Error = 15 / √36
Standard Error = 15 / 6
Standard Error = 2.5
Step 2: Convert the sample mean ages of 98 and 100 months to z-scores.
To find the z-scores, we can use the formula:
z = (x - μ) / standard error
For 98 months:
z1 = (98 - 96) / 2.5
z1 = 0.8
For 100 months:
z2 = (100 - 96) / 2.5
z2 = 1.6
Step 3: Find the probability using the z-table.
Using the z-table or a calculator, we can find the probabilities associated with the z-scores of 0.8 and 1.6 separately.
P(z < 0.8) = 0.7881
P(z < 1.6) = 0.9452
Step 4: Calculate the final probability.
To find the probability that the mean age is between 98 and 100 months, we subtract the probability associated with the lower z-score from the probability associated with the higher z-score:
P(0.8 < z < 1.6) = P(z < 1.6) - P(z < 0.8)
P(0.8 < z < 1.6) = 0.9452 - 0.7881
P(0.8 < z < 1.6) = 0.1571
Therefore, the probability that the mean age of a random sample of 36 vehicles is between 98 and 100 months is approximately 0.1571.
To find the probability that the mean of the age of the 36 randomly selected vehicles is between 98 and 100 months, we can use the Central Limit Theorem.
The Central Limit Theorem states that for a large enough sample size, the distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution.
In this case, the sample size is large (36) and the standard deviation of the population (σ) is known (15). We can calculate the standard deviation of the sampling distribution, also known as the standard error (SE), using the formula:
SE = σ / √n
where σ is the population standard deviation and n is the sample size.
So, in this case, SE = 15 / √36 = 15 / 6 = 2.5
Now, we can convert the given range (98 to 100 months) into z-scores using the formula:
z = (x - μ) / SE
where x represents the value in the range, μ represents the population mean (96 months in this case), and SE represents the standard error.
For the lower limit (98 months):
z(lower) = (98 - 96) / 2.5 = 0.8
For the upper limit (100 months):
z(upper) = (100 - 96) / 2.5 = 1.6
Now, we need to find the probability that the z-score falls between 0.8 and 1.6. We can use a standard normal distribution table or a statistical software to find this probability.
Using a standard normal distribution table, the probability of the z-score falling between 0.8 and 1.6 is approximately 0.2061.
Therefore, the probability that the mean of the age of the 36 randomly selected vehicles is between 98 and 100 months is approximately 0.2061.