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.
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?
3 answers
.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)
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)