At the end of the festival the organizers estimated that a family of participants spent an average of $90.00 with a standard deviation of $15.00.

What’s the probability that the mean amount spent will be NO more than $75?

Try z-scores:

z = (x - mean)/sd

From your problem:

x = 75
mean = 90
sd = 15

Check a z-table using the z-score you calculate from the above data. Find your probability in the table.

.3413 or 34.13% .

Halloween Festival the organizers estimated that a family of participants spent in average $45.00 with a standard deviation of $10.00.

If 49 participants (49 = size of the sample) are selected randomly, what’s the likelihood that their mean spent amount will be within $4 of the population mean? (mean +/- 4)

To determine the probability that the mean amount spent will be no more than $75, we need to use the concept of the standard error of the mean (SEM). The SEM measures the variability of the sample means and is calculated by dividing the standard deviation by the square root of the sample size.

In this case, since the standard deviation is $15.00, we need to determine the sample size (number of families) to calculate the SEM. Unfortunately, that information is not provided in the question, so we will assume a sample size of 30 for demonstration purposes.

First, calculate the SEM using the formula: SEM = (standard deviation) / (square root of sample size).

SEM = $15.00 / √30 ≈ $2.74 (rounded to two decimal places).

Next, we need to calculate the z-score, which represents the number of standard deviations a particular value is from the mean.

To find the z-score for the $75.00 amount spent, use the formula: z = (X - μ) / SEM, where X is the desired value and μ is the mean.

z = ($75.00 - $90.00) / $2.74 ≈ -5.47 (rounded to two decimal places).

Now we need to calculate the probability associated with the z-score using a standard normal distribution table or a calculator.

Since we are interested in finding the probability that the mean amount spent will be no more than $75.00, we want to find the area under the curve to the left of the z-score.

Using a standard normal distribution table, a z-score of -5.47 corresponds to a probability of approximately 0.00000008, or 8 in 100 million.

Therefore, the probability that the mean amount spent will be no more than $75.00 is extremely low, approximately 8 in 100 million.