Question

a set of 50 data values has amean of 21 and a variance of 16.
I. find the standard(z) score for a data value = 31
II. Find the probability of a data value > 31

Answer 1

the zscore allows us to fit the nonstandard numbers into a normal distribution of a mean of 0 and sd or 1

Answer 2

first of; lets get the mean 0 and measure from there

Answer 3

21 - 21 = 0; now move 31 as well, 31-21 = 10

Answer 4

now we want to squeeze it to fit; sd = sqrt(variance);

in this case sd = sqrt(16) = 4

4/4 = 1 so divide it all by 4 to squeeze it into an sd of 1

Answer 5

\[zscore=\frac{31-21}{4}\]

Answer 6

\[zscore=\frac{x-\bar x}{\sigma}\]

Answer 7

10/4 = 2.5 right? we need a ztable to find the area that this corresponds to; unless you want to integrate the normla distribution curve ..

Answer 8

lets look at the right to find 2.5, and match it across the top at .00

that intersects at .9938 as the total area under the curve to the left of the zscore

Answer 9

the total area under the curve = 1; so to find the area "greater than" this 31; we simply subtract the area up to our zscore from 1

Answer 10

1 - .9938 = whats left; .0062

Answer 11

the idea tho is to move our data stats to a mean of 0; and divide the spread to equate to a standard deviation of 1

Answer 12

thank you, you explained it so much better than the book.