Question

The grades on the last history exam had a mean of 80%. Assume the population of grades on history exams is known to be normally distributed with a standard deviation of 5%. What percent of students earn a score between 70% and 80%?

Answer 1

A. 0.022
B. 0.477
C. 0.50
D. 0.522

Answer 2

Please help!!

Answer 3

do you have a TI-83 or TI-84 calculator?

Answer 4

no

Answer 5

Can you walk me through it?

Answer 6

ok let me find one

Answer 7

ok so here is one calculator you can use

http://www.mathportal.org/calculators/statistics-calculator/normal-distribution-calculator.php

you type 80 in the \(\Large \mu\) box for the mean
type 5 for the \(\Large \sigma\) box (standard deviation)

then you pick the option that has \(\Large P( \square <X < \square ) \)
typing 70 and 80 in the two boxes (smaller number goes in the left box)

Answer 8

that calculator is really cool because you can click the checkbox "show explanation" and it gives you a nice visual step by step

Answer 9

Ahhh! Thank you so much! I had the formula, put didn't know what the symbols meant.

Answer 10

yeah it's probably a bit odd to see greek symbols being used

\[\Large \mu = \text{mu}\]
the "mu" is basically the greek letter m (well it's like the letter m in the latin alpabet)
think M for Mean

\[\Large \sigma = \text{sigma}\]
"sigma" is closely linked to the letter S
S for Standard deviation

Answer 11

mu and sigma here are the lower case greek forms

Answer 12

Ok, can you help with another similar to this one?

Answer 13

sure

Answer 14

In a study of 250 adults, the mean heart rate was 70 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 12 beats per minute. What is the 99% confidence interval for the mean beats per minute?
A. 68.9-76.3
B. 70-72
C. 61.2-72.8
D. 68-72

Answer 15

what is the z critical value for the 99% confidence interval ?

Answer 16

2.576 I think

Answer 17

very good, since the solution to

P( -k < X < k ) = 0.99

is roughly k = 2.576

Answer 18

the mean heart rate was 70 beats per minute

so xbar = 70

Answer 19

"Assume the population of heart rates is known to be approximately normal with a standard deviation of 12 beats per minute"

sigma = 12

Answer 20

"In a study of 250 adults" ---> n = 250

Answer 21

You will now use the formulas

\[\Large L = \overline{ x} - z^{*} \frac{\sigma}{\sqrt{n}}\]
\[\Large U = \overline{ x} + z^{*} \frac{\sigma}{\sqrt{n}}\]

Answer 22

\[\Large \overline{x} = 70\]
\[\Large \sigma = 12\]
\[\Large z^{*} \approx 2.576\]
\[\Large n = 250\]

Answer 23

to get the confidence interval (L,U)

Answer 24

I got D, 68-72, is that right?

Answer 25

\[\Large L = \overline{ x} - z^{*} \frac{\sigma}{\sqrt{n}}\]
\[\Large L = 70 - 2.576* \frac{12}{\sqrt{250}}\]
\[\Large L = 68.04495\]
\[\Large L = 68\]

--------------------------------------------------
\[\Large U = \overline{ x} + z^{*} \frac{\sigma}{\sqrt{n}}\]
\[\Large U = 70 + 2.576* \frac{12}{\sqrt{250}}\]
\[\Large U = 71.95505\]
\[\Large U = 72\]

--------------------------------------------------

I agree, the confidence interval is roughly (L,U) = (68,72)

Answer 26

Thank you!! You have helped me so much!

Answer 27

I'm glad I could help out