Respuesta :
Answer:
a) [tex] P(X<100) =0.5[/tex] or 50%
b) So then we are 1 deviation above the mean. Within one deviation from the mean we have 68% of the data and on the remaining tails we need to have 16%. So then the cumulative % until 1 deviation above the mean is 16+68%=84% approximately.
c) [tex]z=\frac{140-100}{20}=2[/tex]
So then we are 2 deviations above the mean. Within two deviation from the mean we have 95% of the data and on the remaining tails we need to have 2.5%. So then the cumulative % until 2 deviations above the mean is 2.5+95%=97.5% approximately.
d) [tex] z = \frac{120-100}{20}=1[/tex]
So then we are 1 deviation above the mean. Within one deviation from the mean we have 68% of the data and on the remaining tails we need to have 16%. So then the cumulative % until 1 deviation above the mean is 16+68%=84% approximately.
e) [tex] z = \frac{60-100}{20}=-2[/tex]
Since we have 95% within two deviation from the mean we need to have (100-95)/2% = 1.5% on the tails and for this reason the answer wuld be 1.5%
f) [tex] z = \frac{120-100}{20}=-1[/tex]
Since we have 68% within one deviation from the mean we need to have (100-68)/2% = 16% on the tails and on above one deviation from the mean we need to have 100- 16-68 =16%
Step-by-step explanation:
For this case we have test scores following a normal distribution given by:
[tex] X \sim N (\mu = 100, \sigma=20)[/tex]
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
Part a
For this case we knwo that the mean is 100 and since the normal distribution is symmetrical we have 50% of the data below the mean and the other 50% above the mean.
[tex] P(X<100) =0.5[/tex] or 50%
Part b
For this case we want to find the shaded area in the first figure attached and in order to do this we can use the z score formula given by:
[tex] z= \frac{x-\mu}{\sigma}[/tex]
If we calculate the z score for 120 we gto:
[tex] z = \frac{120-100}{20}=1[/tex]
So then we are 1 deviation above the mean. Within one deviation from the mean we have 68% of the data and on the remaining tails we need to have 16%. So then the cumulative % until 1 deviation above the mean is 16+68%=84% approximately.
Part c
We can see this on the second figure attached.
We can calculate the z score for 140 and we got:
[tex]z=\frac{140-100}{20}=2[/tex]
So then we are 2 deviations above the mean. Within two deviation from the mean we have 95% of the data and on the remaining tails we need to have 2.5%. So then the cumulative % until 2 deviations above the mean is 2.5+95%=97.5% approximately.
Part d
We can see the plot on the third figure attached.
We can calculate the z score for 80 and we got:
[tex] z =\frac{80-100}{20}=-1[/tex]
Since within one deviation from the mean we have 68% of the data then on the tails we need to have (100-68)/2= 16% so then the answr for this case would be 16%.
Part e
We can see the picture on the third figure attaches.
We can calculate the z score for 60 and we got:
[tex] z = \frac{60-100}{20}=-2[/tex]
Since we have 95% within two deviation from the mean we need to have (100-95)/2% = 1.5% on the tails and for this reason the answer wuld be 1.5%
Part f
We can see the picture on the last figure attaches.
We can calculate the z score for 60 and we got:
[tex] z = \frac{120-100}{20}=-1[/tex]
Since we have 68% within one deviation from the mean we need to have (100-68)/2% = 16% on the tails and on above one deviation from the mean we need to have 100- 16-68 =16%
