n a recent awards​ ceremony, the age of the winner for best actor was 35 and the age of the winner for best actress was 48. For all best​ actors, the mean age is 48.7 years and the standard deviation is 8.9 years. For all best​ actresses, the mean age is 34.7 years and the standard deviation is 11.7 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.

Respuesta :

Answer:

The best actor's age is farther from the mean, so he has the more extreme age when winning the​ award

Step-by-step explanation:

Z-score:

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Whichever z-score's has the highest absolute value, that is, is farther from the mean, has the most extreme age.

Best actor:

Age of 35, so [tex]X = 35[/tex].

For all best​ actors, the mean age is 48.7 years and the standard deviation is 8.9 years, so [tex]\mu = 48.7, \sigma = 8.9[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{35 - 48.7}{8.9}[/tex]

[tex]Z = -1.54[/tex]

Best actrees:

Age of 48, so [tex]X = 48[/tex]

For all best​ actresses, the mean age is 34.7 years and the standard deviation is 11.7 years, so [tex]\mu = 34.7, \sigma = 11.7[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{48 - 34.7}{11.7}[/tex]

[tex]Z = 1.14[/tex]

The best actor's age is farther from the mean, so he has the more extreme age when winning the​ award

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