Use z scores to compare the given values. In a recent awards ceremony, the age of the winner for best actor was 34 and the age of the winner for best actress was 62. For all best actors, the mean age is 43.4 years and the standard deviation is 8.8 years. For all best actresses, the mean age is 38.2 years and the standard deviation is 12.6 years. (All ages are determined at the time of the awards ceremony.) Relative to their genders, who had the more extreme age when winning the award, the actor or the actress? Explain.

We are given that for all the best actors, the mean age is 43.4 years and the standard deviation is 8.8 years. For all best actresses, the mean age is 38.2 years and the standard deviation is 12.6 years.

To find who had the more extreme age when winning the award, the actor or the actress, we will use the z-score method.

Finding the z-score for the actor;

Let X = age of the winner for best actor

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean age = 43.4 years

[tex]\sigma[/tex] = standard deviation = 8.8 years

It is stated that the age of the winner for best actor was 34, so;

z-score for 34 = [tex]\frac{X-\mu}{\sigma}[/tex]

= [tex]\frac{34-43.4}{8.8}[/tex] = -1.068

Finding the z-score for the actress;

Let Y = age of the winner for best actress

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{Y-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = mean age = 38.2 years

[tex]\sigma[/tex] = standard deviation = 12.6 years

It is stated that the age of the winner for best actress was 62, so;

z-score for 62 = [tex]\frac{Y-\mu}{\sigma}[/tex]

= [tex]\frac{62-38.2}{12.6}[/tex] = 1.889

Since the z-score for the actress is more which means that the actress had more extreme age when winning the award.