Question

who can help with statistics

Answer 1

are you using a z table for this or just the 68% 95% 99% approximation technique?

Answer 2

z table

Answer 3

can you explain e ?

Answer 4

one sec

Answer 5

ok

Answer 6

so we know the following

\[z = \frac{Y-\mu}{\sigma}\]

we have the following
\[z1 = \frac{Y1-6.7}{7}\]
and
\[z2 = \frac{Y2-6.7}{7}\]

We want to find P(scores between Y1 and Y2 approximately 95%) = 1-A(z1)-A(z2)

Answer 7

so what are the y1 and y2?

Answer 8

y1 and y2 are the values between which lies 95% of the students scores.

also, since it only says approximately, we know that the approximation is

\[95 percent= \mu\pm2*\sigma \]

Answer 9

so the upper bound might be 67+2(7) = 67+14 = 81

the lower bound might be 67-2(7) = 67-14 = 53

this is an approximation, but since it says approximately, then i think it might work.

Answer 10

i got -2 and 2

Answer 11

how?

Answer 12

use that formula

Answer 13

wait , how did you get that 2 from the formula?

Answer 14

two standard deviations from the mean on each side is usually approximated to mean 95% of the values in a normal distribution.

Answer 15

oh ` i see

Answer 16

for the f i got 79.59, is it right?

Answer 17

i got 73.03.

\\ if the average was greater than the score you were looking at then it would be >50%

i did this

\[\frac{Y-\mu}{\sigma}=\frac{74.29-\mu}{6.3}=.20\]
solve for mu

Answer 18

what about g ?

Answer 19

ok, so mu in f was 73.03

I gotta go, but satellite73 might be able to help you further.

Answer 20

can anyone explain g ?

Answer 21

Let me take a look at it real quick and see if I can help you

Answer 22

ok thanks

Answer 23

just need explaination about g

Answer 24

are you there?

Answer 25

Ok, so you know that 42.2 is 3%, so you need to look up .0300 on your z-score table and that should give you -1.885. Once you find that data value, you use the following formula:

\[z=\frac{ x-\mu }{ \sigma }\]
Where x is the data value, mu is the mean and, sigma is the standard deviation. So, since you know the z-score, you need to solve for the standard deviation

\[-1.885=\frac{ 42.2-61}{ sigma }\]

so
\[sigma=\frac{ 42.2-61 }{ -1.885 }\]

\[sigma=\frac{ -18.8 }{ -1.885 }\]

\[sigma = 9.97\]

>>sorry, it would let me edit the sigma, it should be shown like the first equation- this is the standard deviation

Answer 26

wow ` awesome, for the f i got 79.59

Answer 27

is it rgiht?

Answer 28

am I right?

Answer 29

let me double check my work really quick...

Answer 30

ok

Answer 31

I'm not sure if what I am doing is correct, but my thinking was that if 20% got higher than 74.92 then that point would be 80%. So I looked for the z-score, and got 0.845, then I plugged it into the same equation as above, but solved for the mean instead.

Answer 32

Let me type out the equations for you.

Answer 33

what did you get?

Answer 34

because 20% is 0.2 , so if i look at the z-score, i got -1.4

Answer 35

sorry , i did it wrong

Answer 36

\[z=\frac{ x-\mu }{ \sigma }\]

\[.845=\frac{ 74.29-\mu }{ 6.3 }\]

\[\mu= 74.29-5.3235\]

\[\mu= 68.967\]

Answer 37

how did you get that .845

Answer 38

that is the z-score for 80%

Answer 39

why we look up 80%?

Answer 40

Since it said that 20% of students got more than 74.29, that means the rest of them had to have gotten a value below that, so on a bell curve, 74.29 would be located at the 80% point

Answer 41

oh isee, ` for the last one, i think the biology is right one