The average electric bill in a residential area is $72 for the month of May. The standard deviation is $6. If the amounts of electric bills are normally distributed, find the probability that the mean of the bill for 15 residents will be less thatn $75

You need to work out the standard deviation of the mean of the bill for 15 residents. That's $6//sqrt(15), which is 1.549. So the distribution of the mean of the bill for 15 residents will be Normal, with mean $72 and standard deviation 1.549. You have to work out the probability that the mean will be less than $75, so you need work out how many standard deviations $75 is above $72, which is ($75 - $72) divided by $1.549, which is 1.936. You now need to know what the area to the left of that point (because you want the probability of the mean bill being less that $75) is, and you can get that from a set of Normal probability tables. If you were looking up 1.96 you would get an answer of 0.975, so you know it's going to be a bit less than that. I've just looked it up myself, and I get 0.9735 - so I reckon that's your answer - i.e. just over 97%.

To find the probability that the mean of the bill for 15 residents will be less than $75, you need to use the sampling distribution of the sample means.

The mean of the sampling distribution of the sample means is equal to the population mean, which is $72.

The standard deviation of the sampling distribution of the sample means, also known as the standard error, can be calculated using the formula:

Standard Error = (Standard Deviation) / √n

where n is the sample size.

In this case, the standard deviation is $6 and the sample size is 15:

Standard Error = 6 / √15

Now, we can use the z-score formula to find the probability:

z = (x - μ) / SE

where x is the desired value ($75), μ is the population mean ($72), and SE is the standard error.

z = (75 - 72) / (6 / √15) = 3 / (6 / √15) = √15 / 2

Using a standard normal distribution table or a calculator, we can find the probability associated with this z-score.

Let's calculate it together:

To find the probability that the mean of the bill for 15 residents will be less than $75, we need to use the concept of the sampling distribution of the sample mean.

Step 1: Find the standard deviation of the sample mean:
The standard deviation of the sample mean (also known as the standard error) can be calculated by dividing the standard deviation of the population (in this case, $6) by the square root of the sample size. So, the standard deviation of the sample mean is:
Standard deviation of the sample mean = Standard deviation of the population / Square root of the sample size
= $6 / √15 ≈ $1.549

Step 2: Convert the given value ($75) into a z-score:
To use the normal distribution table, we need to convert the given value ($75) into a z-score. The z-score formula is given by:
z = (x - μ) / σ
where
x is the given value,
μ is the mean of the population, and
σ is the standard deviation of the sample mean.

In this case, x = $75, μ = $72, and σ (standard deviation of the sample mean) = $1.549. Plugging in the values, we get:
z = ($75 - $72) / $1.549 ≈ 1.939

Step 3: Find the probability using the z-score:
Once we have the z-score, we can use a standard normal distribution table to find the probability associated with that z-score. The z-score represents the number of standard deviations above or below the mean in a standard normal distribution.

Looking up the z-score of 1.939 in the standard normal distribution table, we find that the corresponding probability is approximately 0.9732.

Therefore, the probability that the mean of the bill for 15 residents will be less than $75 is approximately 0.9732 or 97.32%.