at a bank, account balances are nomrally distribued with a mean of $637.52 and standard deviation of $623.16. What is the probability that a simple randon same of 400 accounts has mean that exceeds $1650.00? show me steps and calculations i urgently need help
I think you either have a typo or the problem wanted to be out of reasonable range. Are you sure not $650, not $1650 ???
Anyway here is a method:
Sample Mean also has normal distribution
The mean will have the same mean as the original.
However the sigma = original sigma /sqrt(sample size) = 623.16/20
= 31.15
So
The real question is in a normal distribution with mean = 637.52 and sigma = 31.15 what is the probability of over 1650?
how far above mean? (1650 - 637.52) = 1012.48
how many sigmas above mean?
1012.48 / 31.15 = 32.5 sigmas from mean
That is off any table of normal distribution. It is so improbable that the online normal distribution calculators just return zero.
So, lets see what the probability is of the opposite, the mean lying between 0 and 32.5 sigma. It is of course .5
In other words the probability of the sample being beyond 32.5 sigma is so tiny the calculators will not do it.
NOW
What if you had said $650 ?????
The real question is in a normal distribution with mean = 637.52 and sigma = 31.15 what is the probability of over 650?
how far above mean? (650 - 637.52) = 12.48
how many sigmas above mean?
12.48 / 31.15 = .401 sigmas from mean
So from table, what is probability beyond .401 sigma above mean ?
from normal distribution table F(.400) = .345
To solve this problem, we will use the Central Limit Theorem and convert the given situation into a standardized form. Here are the steps to find the probability:
Step 1: Determine the sample size (n), which is given as 400 in this case.
Step 2: Calculate the mean of the sample distribution using the mean of individual account balances: sample mean = population mean = $637.52.
Step 3: Calculate the standard deviation of the sample distribution using the formula: sample standard deviation = population standard deviation / square root of sample size.
sample standard deviation = $623.16 / sqrt(400) = $623.16 / 20 = $31.16.
Step 4: Apply the Central Limit Theorem to standardize the distribution.
z = (required mean - sample mean) / (sample standard deviation / sqrt(n))
z = ($1650 - $637.52) / ($31.16)
z = $1012.48 / $31.16
z ≈ 32.51
Step 5: Find the probability using the standardized z-score.
P(X > $1650) = P(Z > 32.51)
Step 6: Look up the probability in a z-table or use a calculator to find the area under the standard normal curve.
Since the z value of 32.51 is extremely high and likely beyond the range of a standard normal distribution table, we can approximate the result as a very small probability that is almost close to zero.
To summarize, the probability that a simple random sample of 400 accounts has a mean exceeding $1650.00 is very close to zero.