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sswaqqin
14.02.2020 •
Mathematics
Find the z-score corresponding to the given area. Remember, z is distributed as the standard normal distribution with mean µ = 0 and standard deviation σ = 1. (12 pts.)a. The area to the left of z is 15%.b. The area to the right of z is 65%.c. The area to the left of z is 10%.d. The area to the right of z is 5%e. The area between –z and z is 95%. (Hint: draw a picture and figure out the area to the left of the –z)f. The area between –z and z is 99%
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Ответ:
a)![z_{\alpha}= -1.036](/tpl/images/0511/1816/bcb12.png)
b)![z_{\alpha}= -0.385](/tpl/images/0511/1816/e4b47.png)
c)![z_{\alpha}= -1.282](/tpl/images/0511/1816/24e33.png)
d)![z_{\alpha}= 1.645](/tpl/images/0511/1816/448f0.png)
e)![z_{\alpha/2}= -1.960 , z_{\alpha/2}= 1.960](/tpl/images/0511/1816/e10e2.png)
f)![z_{\alpha/2}= -2.576 , z_{\alpha/2}= 2.576](/tpl/images/0511/1816/6be69.png)
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Part a
Let Z the random variable that represent the variable of a population, and for this case we know the distribution for Z is given by:
Where
and ![\sigma=1](/tpl/images/0511/1816/99a37.png)
We want 15% of the area in the left so we can use z table or the following code in excel in order to find the value required:
"=NORM.INV(0.15,0,1)"
And we got![z_{\alpha}= -1.036](/tpl/images/0511/1816/bcb12.png)
Part b
We want 65% of the area in the right and that implies 100-65=35% of the area in the left so we can use z table or the following code in excel in order to find the value required:
"=NORM.INV(0.35,0,1)"
And we got![z_{\alpha}= -0.385](/tpl/images/0511/1816/e4b47.png)
Part c
We want 10% of the area in the left so we can use z table or the following code in excel in order to find the value required:
"=NORM.INV(0.10,0,1)"
And we got![z_{\alpha}= -1.282](/tpl/images/0511/1816/24e33.png)
Part d
We want 5% of the area in the right and that implies 100-5=95% of the area in the left so we can use z table or the following code in excel in order to find the value required:
"=NORM.INV(0.95,0,1)"
And we got![z_{\alpha}= 1.645](/tpl/images/0511/1816/448f0.png)
Part e
For this case we waant two values that accumulates 95% of the area in the middle. That means we need to have 100-95 % = 5% of the area in the tails and since the distribution is symmetric we need to have 5/2 = 2.5% of the area on each tail. So we can use the following codes to find the two values:
"=NORM.INV(0.025,0,1)"
"=NORM.INV(1-0.025,0,1)"
Part f
For this case we waant two values that accumulates 99% of the area in the middle. That means we need to have 100-99 % = 1% of the area in the tails and since the distribution is symmetric we need to have 1/2 = 0.5% of the area on each tail. So we can use the following codes to find the two values:
"=NORM.INV(0.005,0,1)"
"=NORM.INV(1-0.005,0,1)"
Ответ:
-6.2-(-3.96) = -2.24
-3.96-(-6.2)= -2.24
5.71+2.84=8.55
2.84+5.71=8.55
Hope this helped :)