Find the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean 90 and a standard deviation 60 is larger than 100
To find the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean of 90 and a standard deviation of 60 is larger than 100, we can use the Central Limit Theorem.
According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the distribution of the original population, as long as the sample size is sufficiently large.
In this case, the sample size is 25, which is considered large enough for the Central Limit Theorem to be applicable.
To calculate the probability, we need to standardize the sample mean using the formula:
Z = (X - μ) / (σ / sqrt(n))
where:
Z = the standardized z-score
X = the desired value of the sample mean (100 in this case)
μ = the mean of the population (90)
σ = the standard deviation of the population (60)
n = sample size (25)
Now let's plug in the values:
Z = (100 - 90) / (60 / sqrt(25))
Simplifying:
Z = 10 / (60 / 5)
Z = 0.8333
Now, we can look up the probability associated with the standardized z-score of 0.8333 in the standard normal distribution table.
Using the z-table or a calculator, we find that the probability is approximately 0.7967.
So, the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean of 90 and a standard deviation of 60 is larger than 100 is 0.7967, or 79.67%.
To find the probability that the mean of a random sample is larger than 100, we need to use the Central Limit Theorem (CLT) and the properties of the normal distribution.
Step 1: Understand the problem
We are dealing with a normally distributed population with a mean (μ) of 90 and a standard deviation (σ) of 60. We are taking a random sample of 25 elements from this population and want to find the probability that the mean of this sample is larger than 100.
Step 2: Apply the Central Limit Theorem
The Central Limit Theorem tells us that regardless of the shape of the population distribution, if we take a large enough random sample, the distribution of the sample means will approach a normal distribution. In this case, since we are taking a sample size of 25, we can assume that the sample means will be approximately normally distributed.
Step 3: Calculate the z-score
To use the standard normal distribution, which has a mean of 0 and a standard deviation of 1, we need to calculate the z-score for the value 100. The z-score formula is given by:
z = (x - μ) / (σ / √n)
where x is the value we are interested in, μ is the mean of the population, σ is the standard deviation of the population, and n is the sample size.
In our case, x = 100, μ = 90, σ = 60, and n = 25. Plugging these values into the formula, we get:
z = (100 - 90) / (60 / √25) = 10 / 12 = 0.8333
Step 4: Find the probability using the standard normal distribution
Now that we have the z-score, we can use a standard normal distribution table or a calculator to find the probability.
The probability (P) that the sample mean is larger than 100 can be found by calculating the area under the standard normal curve to the right of the z-score of 0.8333. This represents the proportion of the distribution that is greater than 100.
Using a standard normal distribution table or a calculator, we find that the probability is approximately 0.2031.
Therefore, the probability that the mean of a random sample of 25 elements from a normally distributed population with a mean of 90 and a standard deviation of 60 is larger than 100 is approximately 0.2031, or 20.31%.
Note: The result is an approximation because we are using the Central Limit Theorem and assuming that the sample size is large enough for the distribution of the sample means to be approximately normal.
David Lane normal distribution