The weekly earnings of students in one age group are normally distributed with a standard deviation of $110. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 99 percent confidence that the sample mean will not differ from the population mean by more than $10.

806

Step-by-step explanation:

Given that:

Standard deviation (s) = $110

Error = $10

Zcritical ; Zα/2 at 99% confidence interval = 2.58

Using the relation :

[(Zα/2 * s) / sqrt(n)] < 10

[(2.58 * 110) /sqrt(n)] = 10

283.8 / sqrt(n) = 10

283.8 = 10 * sqrt(n)

Sqrt(n) = 283.8/ 10

Sqrt(n) = 28.38

Square both sides

n = 28.38^2

n = 805.4244

n = 806

Hence, sample size = 806

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: