Answer:
a) From the mean and 2 deviations above we have 95/2% = 47.5% of the data
b) Below one deviation from the mean we have (50-34)% =16% of the data
c) One deviation below the mean and 2 deviations above we have (47.5%) +34% = 81.5% of the data
Step-by-step explanation:
For this case we know that for a random variable X we have:
[tex] X \sim N(\mu = 34, \sigma =2.5)[/tex]
The empirical rule states that within one deviation from the mean we have 68% of the values, within 2 deviations 95% and ithin 3 deviations 99.7% of the data.
So then we want to find the % between 34 and 39. We can use the z score formula given by:
[tex] z =\frac{X-\mu}{\sigma}[/tex]
In order to calculate the deviations from the mean and replacing we got:
[tex] z = \frac{34-34}{2.5}=0[/tex]
[tex] z = \frac{39-34}{2.5}=2[/tex]
And from the mean and 2 deviations above we have 95/2% = 47.5% of the data
For the probability that would be less than 31.5 we find the z score:
[tex] z = \frac{31.5-34}{2.5}=-1[/tex]
And below one deviation from the mean we have (50-34)% =16% of the data
And for the last part between 29 and 36.5 we find the z scores and we got:
[tex] z = \frac{29-34}{2.5}=-2[/tex]
[tex] z = \frac{36.5-34}{2.5}=1[/tex]
And from one deviation below the mean and 2 deviations above we have (47.5%) +34% = 81.5% of the data
