Statistics: normal distribution

In summary: But thank you both for the help!In summary, the problem involved finding the probability that a normally distributed variable with mean 95 and standard deviation 17 would be greater than 126.79. After calculating the z-score and using a statistical table, the probability was found to be 0.0307. For the second part of the problem, the value of X that only the top 16% exceeded was found by using the fact that the top 16% must be equal to 84% of the total distribution. This resulted in a z-score of 1.0, which was then solved for X to be 112.
  • #1
k77i
28
0

Homework Statement



Suppose that X is normally distributed with mean 95 and standard deviation 17.

A. What is the probability that is greater than 126.79?

B. What value of X does only the top 16% exceed?



Homework Equations



z= (X-(mean of x))/standard deviation



The Attempt at a Solution



using the equation, I found z= (126.79-95)/17=1.87

then using a statistical table, the value for P(z greater than 1.87) = 1 - P(z=1.87) = 0.0307

But this is wrong..

For B)

I considered that 84% must be the value which the top 16% must exceed.. then 84% of 95 is 88.42

Also wrong (well I wasn't really too sure about part B)

Any help would be appreciated!
 
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  • #2
The tabular value for above that z-score seems correct. Your z-score seems correct to the given number of decimal places. How do you know that the answer under a) is "wrong"?

For B), you should find such a [itex]Z_{0}[/itex] that:

[tex]
P(Z > Z_{0}) = 0.16 \Rightarrow P(Z < Z_{0}) = 0.84
[/tex]

Using that value for [itex]Z_{0}[/itex], solve for X in the expression:
[tex]
Z = \frac{X - \mu}{\sigma}
[/tex]
where [itex]\mu[/itex] is the mean and [itex]\sigma[/itex] is the standard deviation.
 
  • #3
k77i said:

Homework Statement



Suppose that X is normally distributed with mean 95 and standard deviation 17.

A. What is the probability that is greater than 126.79?

B. What value of X does only the top 16% exceed?



Homework Equations



z= (X-(mean of x))/standard deviation



The Attempt at a Solution



using the equation, I found z= (126.79-95)/17=1.87

then using a statistical table, the value for P(z greater than 1.87) = 1 - P(z=1.87) = 0.0307

But this is wrong..
No, that is correct. Why do you say it is wrong?

For B)

I considered that 84% must be the value which the top 16% must exceed.. then 84% of 95 is 88.42

Also wrong (well I wasn't really too sure about part B)

Any help would be appreciated!
95 is the mean- 50% are larger than 95!

Looking at a table of the normal distribution, N(x< a)= 0.84 for a= 1.0 (approximately).

[tex]\frac{z- 95}{17}= 1.0[/tex]
solve for z.
 
  • #4
Thank you both for the help with part B. As for part A, since its a webwork assignment, it tells me when I enter an incorrect answer. I guess it's possible that there might be some sort of problem with the webwork.
 
  • #5
maybe you need to use the correct number of significant figures.
 
  • #6
I put all the numbers (0.0307).. The chart I have for the z-score only goes upto 4 decimal places so I'm not sure how much it's actually supposed to go up to
 
  • #7
Well, assuming the mean and the standard deviation are known to the same number of significant figures as 126.79, then the z-score is:
[tex]
\frac{126.79 - 95}{17} = \frac{31.79}{17} = 1.870
[/tex]
to 4 significant figures (actually the quotient is exact). For this, the probability in question is:

1 - 0.9693 = 0.0307

So, it can't be that.

Are you sure this is the standard deviation and not the variance?
 
  • #8
Yes it is the standard deviation. I copied the question word for word..
In any case it's too late, the homework set is closed and apparently the accepted answer was 0307419079428789. I think I was supposed to use excel to find the exact value to the correct amount of significant figures..
 
  • #9
lol.
 

FAQ: Statistics: normal distribution

1. What is a normal distribution?

A normal distribution, also known as a Gaussian distribution, is a statistical concept that describes a symmetrical bell-shaped curve. In a normal distribution, the majority of values fall close to the mean, with fewer values on the outer ends of the curve.

2. What are the characteristics of a normal distribution?

A normal distribution is characterized by three main properties: symmetry, unimodality (having a single peak), and the 68-95-99.7 rule. This rule states that approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

3. How is a normal distribution used in statistics?

A normal distribution is used in statistics to model many natural phenomena, such as human height, IQ scores, and blood pressure. It is also commonly used in hypothesis testing, as many statistical tests assume that the data follows a normal distribution.

4. What is the significance of the mean and standard deviation in a normal distribution?

The mean and standard deviation are important measures in a normal distribution. The mean represents the center of the distribution, while the standard deviation measures the spread of the data around the mean. These values are used to calculate the probabilities of different outcomes and to make comparisons between different sets of data.

5. How do you determine if a dataset follows a normal distribution?

There are a few ways to determine if a dataset follows a normal distribution. One way is to visually inspect the data using a histogram or a normal probability plot. Another way is to use statistical tests, such as the Shapiro-Wilk test or the Kolmogorov-Smirnov test, which compare the data to a theoretical normal distribution. If the p-value from these tests is greater than 0.05, the data can be considered to follow a normal distribution.

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