What is the z-score of sample X, if $n=196,\text{}\mu =71,\text{}\text{St.Dev}=42,\text{}{\mu}_{X}=83$?

Mauricio Mathis
2022-07-17
Answered

What is the z-score of sample X, if $n=196,\text{}\mu =71,\text{}\text{St.Dev}=42,\text{}{\mu}_{X}=83$?

You can still ask an expert for help

Eve Good

Answered 2022-07-18
Author has **18** answers

Given

$n=196\phantom{\rule{0ex}{0ex}}\mu =71\phantom{\rule{0ex}{0ex}}\text{St.Dev}=42\phantom{\rule{0ex}{0ex}}{\mu}_{X}=83$

Find the Standard Error

$SE=\frac{\sigma}{\sqrt{n}}=\frac{42}{\sqrt{196}}=3$

Find the z-score

$z=\frac{71-83}{3}=-4$

$n=196\phantom{\rule{0ex}{0ex}}\mu =71\phantom{\rule{0ex}{0ex}}\text{St.Dev}=42\phantom{\rule{0ex}{0ex}}{\mu}_{X}=83$

Find the Standard Error

$SE=\frac{\sigma}{\sqrt{n}}=\frac{42}{\sqrt{196}}=3$

Find the z-score

$z=\frac{71-83}{3}=-4$

asked 2022-10-21

A random variable X is normally distributed with $\mu =60$ and $\sigma $ = 3. What is the value of 2 numbers a,b so that $P(X=a)=P(X=b)$.

The solution is $a=60$ and $b=65$.

However, I do not know how to come up with that answer. As far as I understand $P(X=a)$ and $P(X=b)$ have to be both 0 since you always have to give a range e.g. $P(a<X)$. Moreover if I insert the values 60 and 65 in the formula $Z=(X-\mu )/\sigma $ than I would end up with 0,1.667 and z-scores 0.5, 0.952 respectively.

The solution is $a=60$ and $b=65$.

However, I do not know how to come up with that answer. As far as I understand $P(X=a)$ and $P(X=b)$ have to be both 0 since you always have to give a range e.g. $P(a<X)$. Moreover if I insert the values 60 and 65 in the formula $Z=(X-\mu )/\sigma $ than I would end up with 0,1.667 and z-scores 0.5, 0.952 respectively.

asked 2022-08-17

What is the z-score of sample X, if $n=121,\mu =33,St.Dev.=110,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}E\left[X\right]=32$

asked 2022-09-23

In the Z-score formula, can the standard error measurement be used instead of standard deviation?

asked 2022-07-13

I'm curious what the phrase "on average" means. Here is an example:

On average, $30\mathrm{\%}$ were further than ___ kilometers away when they had their accident.

Is $30\mathrm{\%}$ a z-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $\mu =30$ kilometers and $\sigma =8.0$ kilometers.

On average, $30\mathrm{\%}$ were further than ___ kilometers away when they had their accident.

Is $30\mathrm{\%}$ a z-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $\mu =30$ kilometers and $\sigma =8.0$ kilometers.

asked 2022-10-08

What is the z-score of sample X, if $n=256,\text{}\mu =17,\text{}\text{St.Dev}=80,\text{}{\mu}_{X}=22$?

asked 2022-08-22

What is the z-score of sample X, if $n=100,\mu =45,SD=30,\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}E\left[X\right]=31$?

asked 2022-06-25

I have learned that z-scores are "Measures of Relative Standing" and that they can be used to compare two scores that are from different normal distributions.

For example if I wanted to compare how I performed

with a 85% in a math class with mean 70 and std dev 15;

and

with a 70% in a english class with mean 50 and std dev 15;

I would use the z-score formula to calculate a relative value for two situation and use it to compare them.

However why is it not enough just to compare each classes with the mean and use this as a basis for comparison? For example

85% -70% = 15%

70% - 50% = 20%

20 > 15 therefore I must have performed better in english

I know this is incorrect but I don't understand why. thx for the help

For example if I wanted to compare how I performed

with a 85% in a math class with mean 70 and std dev 15;

and

with a 70% in a english class with mean 50 and std dev 15;

I would use the z-score formula to calculate a relative value for two situation and use it to compare them.

However why is it not enough just to compare each classes with the mean and use this as a basis for comparison? For example

85% -70% = 15%

70% - 50% = 20%

20 > 15 therefore I must have performed better in english

I know this is incorrect but I don't understand why. thx for the help