Question

Assume that a set of test scores is normally distributed with a mean of 100 and a standard deviation of 20. Use the 68-95-99.7 rule to find the following quantities:___________
Suggest you make a drawing and label first…
A. Percentage of scores less than 100
B. Relative frequency of scores less than 120
C. Percentage of scores less than 140
D. Percentage of scores less than 80
E. Relative frequency of scores less than 60
F. Percentage of scores greater than 120

Answer 1

Answer:

a) $P(X<100) =0.5$ 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) $z=\frac{140-100}{20}=2$

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) $z = \frac{120-100}{20}=1$

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) $z = \frac{60-100}{20}=-2$

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) $z = \frac{120-100}{20}=-1$

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:

$X \sim N (\mu = 100, \sigma=20)$

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.

$P(X<100) =0.5$ 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:

$z= \frac{x-\mu}{\sigma}$

If we calculate the z score for 120 we gto:

$z = \frac{120-100}{20}=1$

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:

$z=\frac{140-100}{20}=2$

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:

$z =\frac{80-100}{20}=-1$

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:

$z = \frac{60-100}{20}=-2$

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:

$z = \frac{120-100}{20}=-1$

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%