Answer:
a) [tex] 61-35.8=25.3[/tex]
b) [tex] \frac{25.2}{11.3}=2.23[/tex] deviations
c) [tex] z = \frac{61- 35.8}{11.3}= 2.23[/tex]
d) For this case since we have that z>2 we can consider this value as unusual, since is outside of the interval considered usual.
Step-by-step explanation:
Assuming this complete question : "Helen Mirren was 61 when she earned her Oscar-winning Best Actress award. The Oscar-winning Best Actresses have a mean age of 35.8 years and a standard deviation of 11.3 years"
a) What is the difference between Helen Mirren’s age and the mean age?
For this case we can do this:
[tex] 61-35.8=25.2[/tex]
b) How many standard deviations is that?
We just need to take the difference and divide by the deviation and we got:
[tex] \frac{25.2}{11.3}=2.23[/tex] deviations
c) Convert Helen Mirren’s age to a z score.
The z score is defined as:
[tex] z = \frac{x- \mu}{\sigma}[/tex]
And if we replace the values given we got:
[tex] z = \frac{61- 35.8}{11.3}= 2.23[/tex]
d) If we consider “usual” ages to be those that convert to z scores between –2 and 2, is Helen Mirren’s age usual or unusual?
For this case since we have that z>2 we can consider this value as unusual, since is outside of the interval considered usual.