How do you find the area under the normal distribution curve to the right of z = –3.24?

akuzativo617
2022-11-18
Answered

How do you find the area under the normal distribution curve to the right of z = –3.24?

You can still ask an expert for help

gortepap6yb

Answered 2022-11-19
Author has **19** answers

Considering that you have a fixed amount b and a variable one m that depends on the number of copies x sold, you can use the general form for a (linear) equation:

y=mx+b

$C\left(x\right)=0.45x+2,050,000$

y=mx+b

$C\left(x\right)=0.45x+2,050,000$

asked 2022-07-19

A survey found that women's heights are normally distributed with mean 63.4

in and standard deviation 2.4

in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

a) The percentage of women who meet the height requirement is:

The website says I am supposed to use $\frac{z-\lambda}{2.4}$

doing so gives me $\frac{16.6}{2.4}=7.086$ This value literally is useless. What am I doing wrong?

in and standard deviation 2.4

in. A branch of the military requires women's heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

a) The percentage of women who meet the height requirement is:

The website says I am supposed to use $\frac{z-\lambda}{2.4}$

doing so gives me $\frac{16.6}{2.4}=7.086$ This value literally is useless. What am I doing wrong?

asked 2022-07-22

13 % of people took a math course. What is the probability that in a 350 randomly selected sample, less than 40 people take the course?

So I have $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

So I have $X\sim Bin(350,\frac{13}{100})$

Then let $Y\sim N(\frac{13}{100},{\sqrt{0.00032314}}^{2})$

I want $P(X<40)=P(\hat{p}<\frac{4}{10})$

Then $P(z<\frac{\frac{4}{10}-\frac{13}{100}}{{\sqrt{0.00032314}}^{2}})$ which is obviously very wrong.

Where did it go wrong, and why did it go wrong?

asked 2022-08-06

What is the z-score of X, if n = 135,$\mu =74$, SD =3, and X =73?

asked 2022-08-16

What is the z-score of sample X, if $$

asked 2022-08-12

Fine the area , to the nearest thousandth, of the standard distribution between the given z-scores. Z equals 0 and z equals 1.5

asked 2022-08-15

What is the z-score of sample X, if $n=256,\text{}\mu =88,\text{}\text{St.Dev}=128,\text{}{\mu}_{X}=96$?

asked 2022-08-24

What is the z-score of X, if n = 57, $\mu =35$, SD =5, and X =13?