The average height of BYU freshman from a random sample of 450 freshman at BYU is 68 inches with a standard deviation of 1.5 inches. What is the 98% confidence interval for the average height of BYU freshman? a. (59.86, 60.14)
b. (67.83, 68.17)
c. (66.5. 69.5)
d. (67.86, 68.14)

b. (67.83, 68.17)

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 2.327.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 68 - 0.17 = 67.83.

The upper end of the interval is the sample mean added to M. So it is 68 + 0.17 = 68.17.

This means that the correct answer is given by option B.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

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