A population is normally distributed, with a mean of 23.45 and a standard deviation of 3.8. What is the probability of each of taking a sample of size 10 and obtaining a sample mean of 22 or more?
To calculate the probability of obtaining a sample mean of 22 or more, we need to use the central limit theorem. According to the central limit theorem, the sampling distribution of the sample means will approximately follow a normal distribution, regardless of the shape of the population distribution, as long as the sample size is sufficiently large (typically n > 30).
In this case, the sample size is 10, which is not large enough to assume a normal distribution for the sample means. However, we can still approximate the probability using the normal approximation to the binomial distribution.
First, we need to calculate the standard error of the mean (SE) using the formula:
SE = σ / √n
where σ is the standard deviation of the population and n is the sample size.
SE = 3.8 / √10
SE ≈ 1.20
Next, we need to calculate the z-score using the formula:
z = (x - μ) / SE
where x is the sample mean, μ is the population mean, and SE is the standard error.
z = (22 - 23.45) / 1.20
z ≈ -1.20
We want to find the probability of obtaining a sample mean of 22 or more, which is equivalent to finding the area to the right of the z-score -1.20 in the standard normal distribution. This can be calculated using a standard normal distribution table or a statistical software.
Using a standard normal distribution table or a statistical software, we find that the probability of obtaining a sample mean of 22 or more is approximately 0.8869, or 88.69%.
Therefore, the probability of taking a sample of size 10 and obtaining a sample mean of 22 or more is approximately 0.8869, or 88.69%.
To find the probability of obtaining a sample mean of 22 or more from a normally distributed population, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, as long as the sample size is sufficiently large.
To calculate the probability, we can use the z-score formula:
z = (x - μ) / (σ / √n)
Where:
z is the z-score,
x is the sample mean,
μ is the population mean,
σ is the population standard deviation,
n is the sample size.
In this case:
x = 22 (sample mean we are interested in)
μ = 23.45 (population mean)
σ = 3.8 (population standard deviation)
n = 10 (sample size)
First, let's calculate the z-score:
z = (22 - 23.45) / (3.8 / √10)
z = -1.45 / (3.8 / √10)
z ≈ -1.45 / 1.1998
z ≈ -1.209
Next, we need to find the area to the right of the z-score on the standard normal distribution. We can use a z-table or a statistical calculator to find this probability.
Using a z-table or calculator, we find that the area to the right of z = -1.209 is approximately 0.886
Therefore, the probability of obtaining a sample mean of 22 or more is approximately 0.886 or 88.6%.