A population has a mean of 200 and a standard deviation of 50. Suppose a simple random sample of size 100 is selected and is used to estimate μ.
What is the probability that the sample mean will be within� 5 of the population mean?
To find the probability that the sample mean will be within 5 of the population mean, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution, as long as the sample size is large enough (n > 30).
In this case, we have a simple random sample of size 100. The sample mean will also have a mean of 200, which is the same as the population mean. However, the standard deviation of the sample mean will be the population standard deviation divided by the square root of the sample size.
Standard deviation of sample mean = population standard deviation / square root of sample size
= 50 / √100
= 50 / 10
= 5
Now that we have the standard deviation of the sample mean, we can use the normal distribution to find the probability that the sample mean will be within 5 of the population mean.
To do this, we need to find the area under the normal curve between the population mean minus 5 and the population mean plus 5. This can be done by finding the z-scores corresponding to these values and then using a standard normal distribution table or calculator.
The z-score is calculated using the formula:
z = (x - μ) / σ
Where:
x = the value we are interested in (population mean plus or minus 5)
μ = population mean
σ = standard deviation of sample mean
For the population mean plus 5:
z1 = (μ + 5 - μ) / 5 = 5 / 5 = 1
For the population mean minus 5:
z2 = (μ - 5 - μ) / 5 = -5 / 5 = -1
Using a standard normal distribution table or calculator, we can find the area between z1 and z2. The area between z1 and z2 represents the probability that the sample mean will be within 5 of the population mean.
By looking up the z-scores in a standard normal distribution table or using a calculator, we find:
The area to the left of z1 (1) is approximately 0.8413.
The area to the left of z2 (-1) is approximately 0.1587.
Therefore, the probability that the sample mean will be within 5 of the population mean is:
0.8413 - 0.1587 = 0.6826
So, there is approximately a 68.26% probability that the sample mean will be within 5 of the population mean.