Question

Question is: find 60th percentile of distribution with mu = 266 and sigma = 16. Our instructor told us P=(x < .25) = .60. In other words find Q3. I don't get this. Q3 corresponds to 75th percentile.

Answer 1

is it a normal distribution?

Answer 2

if its normal, then you want to find a zscore related to 60%

Answer 3

something like 0.25

Answer 4

\[z=\frac{x-mean}{sd}\]
\[z(sd)=x-mean\]
\[z(sd)+mean=x\]
\[z_{.6000}~(16)+266 =x\]

Answer 5

Z .6000 is about .25, which gives us:

.25(16) + 266 = 270

Answer 6

im curious why you thing this is Q3 related?

Answer 7

My instructor said, you're basically looking for Q 3. I don't get it either but this is the problem: P(x < .25) = .60. Use standard normal distribution table to find z/zscore. Z score 0.25 is between .5987 and .6026. solve by: P(.25) (16) + 266= 270. 270 < .25. Looking at my notes, I just don't get ow he came up with .25....
I'm also looking for a tutor to work with personally...any info appreciate

Answer 8

the notation is a little off: P(Z < .25) = P(X < x), you are trying to determine "x"

Answer 9

Your instructor prolly mean that it is similar to finding Q3

Answer 10

finding a percentile and finding a quartile are relatively the same process

Answer 11

.25 is from the ztable

look for a field value closes to .6000, or in some books it may be .1000

Answer 12

youre field values are

.5987 and .6026. ; the average between these is .6006

.5978 .6006 .6026
^^
.6000

since we are closer to .5978, use its zscore

add the row and column value that it is in together

Answer 13

should be: 0.2 + .05 = 0.25 for a zscore

Answer 14

Oh, my goodness! You're probably right that he meant to say "it's like finding Q 3" but I sometimes miss points he makes when I'm writing and listening....THANK YOU amistre64...Are you a teacher, statastician...

Answer 15

im some old guy that gets bored at work ... so maybe lol