Question

Scores on a standardized test are normally distributed with a mean of 350 and a standard deviation of 25. Approximately what percent of students scored between 360 and 380?
10%

16%

23%

38%



Question 9.9. In a standard normal distribution, above which z-score do approximately 57% of the data lie?
1.72

0.18

0.18

1.72

Answer 1

@Kitten_is_back

Answer 2

@chlobohoe

Answer 3

which one do u think is your andwer

Answer 4

okay so do you know the formula for this babe?

Answer 5

I was thinking number 1 was C and number 2 is D

Answer 6

im pretty sure youre right about number 1 being C.. do you know how you got that?

Answer 7

@chlobohoe I'm sorry I really don't know I took a guess for it

Answer 8

okay so we know two things..
μ = 350
σ = 25
dont let the fancy letters freak you out tho

Answer 9

the formula should look something like this..
\[z=\frac{ x-\mu }{ \sigma } \]

Answer 10

okay yes 350 is the mean and 25 is the standard deviation

Answer 11

okay how do I find the percent to that

Answer 12

My answer is 14.4,15.2

Answer 13

standardize x to z = (x - μ) / σ
P( 360 < x < 380) = P[( 360 - 350) / 25 < Z < ( 380 - 350) / 25]
P( 0.4 < Z < 1.2) = P( z < 1.2) - P( z < 0.4) = 0.8849 - 0.6554 = 0.2295
= 23 %

Answer 14

and if ur wondering how we changed the original equation to what we did
z(360) = 10/25 = .4
z(380) = 30/25 = 1.2
P(350 < x < 380) = normalcdf(.4, 1.2)

Answer 15

Hmmmm okay I think I got it a little

Answer 16

Thank you