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i want $10000000000
Try and post yr replies on the earlier question, otherwise they won't make sense without much context :).
A student bought five books for the Fall semester. The costs of the books were $52, $70, $73, $63, and $47. Determine the student's average book cost? Determine the standard deviation of the students books? Determine the median price of the students books?
is it $61? Is it $11.25? and is it $63?
does anyone know if this is the correct answers?
305/5 = 61
it may be 61. or not. How does one know?
thank you amistre64
do you know I find the standard deviation?
...i think i recall it; how many numbers?
52, 70, 73, 63, 47 (5)
whats the (5) doing there?
total of five books, but you confirmed my answer, thank you.
did you hear of statcrunch?
47-63 = -16 52-63 = -9 63 mean= 0 70-63 = 7 73-63 = 10 squre the results
std deviation is 9.2 i think
256 81 49 100 ---- 486
divide by 5-1? i think now
I got the standard deviation at $11.25, is that incorrect?
486 /4 ---- 121.5
square root that?
oh you may be right, I used the incorrect formula for std deviation
11.02 or so....
I got 11.247, so I rounded it to 11.24?
I meant 11.25
11.25 is close :)
i think i did the steps right lol
i know what i did wrong, i used the median and not the mean lol
A researcher investigated last month's long distance telephone charges for residents of a small city. Ron lives in the city, and the Z-score of his last month's long distance telephone charges was -2.8. Describe Ron's charges as they relate to the average city resident?
Do I use a normal distribution table to find Z?
Ron’s charges represent only .3% of the residents who utilize long distance telephone services. (z-score: 0.0025)
47-61 = 14 196 52-61 = 9 81 63-61 = 2 4 70-61 = 9 81 73-61 = 12 144 ---- 506/4 = 126.5 sqrt(126.5) = 11.247 got it finally lol
thank you, your awesome!
A researcher investigated last month's long distance telephone charges for residents of a small city. Ron lives in the city, and the Z-score of his last month's long distance telephone charges was -2.8. Describe Ron's charges as they relate to the average city resident? Ron’s charges represent only .3% of the residents who utilize long distance telephone services. (z-score: 0.0025)
i havent heard of a z score before...new terminology for me ..
do you statistics?
taking stats and calc1 over the summer
taught myself some of it so far, but stats has been a backburner subject for me :)
I know it is difficult, but not impossible