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I am trying to figure out the formula in using the STDEV and Standard Error with the mean to find the 95% confidence interval. I have all the information but cannot seem to find the correct formula. Any help would be appreciated

Solved by F. H. in 30 mins

What does this mean? =COUNTIF(M3:M31,4)
my understanding is the m3 to m31 but im not sure what the ,4 means?

Solved by K. J. in 21 mins

CRM cannot automate
Exported to excel to generate payment dates
I have two formulas
Paid 1 month after start date
Paid 1 month after date sold
This mean i must create two separate tabs then combine them so i can upload back into CRM
Are you able to merge the code so i can simply copy the code then drag down then upload?

Solved by I. L. in 19 mins

Here's the question, ''for participants estimates for the jar with 100 jelly beans calculate and present the mean, median, and standard deviation. In addition, calculate the proportion of estimates that were closer than the mean to the correct answer.'' I've already calculated the mean, median and standard deviation but don't know how to do the 2nd part of the question, which is the proportion part.

Solved by B. H. in 25 mins

Question: calculate and present the mean and standard deviations for the two variables representing the proportions for knowledge questions and proportions for jelly bean estimates. Use a t-test and report the p-level.

Solved by M. D. in 19 mins

I have a formula A8*C$2
What does that mean?
What does $ mean?

Solved by T. D. in 23 mins

Hi I'm having trouble with Binomial and Trinomial trees in excel. I need to find the mean and standard deviation for them using factorial formulas, but that is where my understanding of excel is letting me down.

Solved by M. F. in 12 mins

use the excel sheet name part 2 to answer the question and round the answer to 4 decimal places . Information from the American institute of insurance indicates that mean amount of life insurance per household in the united states is $100,000 with a standard deviation of $40,000 assume the population distribution is normal. A random sample of 100 household is taken
What is the probability that sample mean will be more than $120,000?
What is the probability that sample mean will be between $100,000 and $120,000?

Solved by B. Y. in 21 mins

Information from the American institute of insurance indicates the mean amount of life insurance per household in the United States is $110,000 with a standard deviation of $40,000. Assume the population distribution is normal. A random sample of 100 households is taken.
(a) What is the probability that sample mean will be more than $120,00?
(b) What is the probability that sample mean will be between $100,000 and $120,000?
Round your answer to 4 decimal places. ( use part 2 sheet to answer this question)

Solved by F. Y. in 19 mins

The file name sheet part two is kept blank, use part 2 sheet to answer this question. Put a box around your answers. Round your answer to 4 decimal places.
Information from the American institute of insurance indicates the mean amount of life insurance per household in the unites states is $110,00 with a standard deviation of $40,000. Assume the population distribution is normal. A random sample of 100 households is taken.
A. what is the probability that sample mean will be more than $120,000?
B. what is the probability that sample mean will be between $100,00 and $120,000?

Solved by C. F. in 29 mins