The average age of doctors in a certain hospital is 48.0 years old. suppose the distribution of ages is normal and has a standard deviation of 6.0 years. if 9 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 48.8 years. assume that the variable is normally distributed.

Question

Sarge

Mathematics

26 July, 08:36

The average age of doctors in a certain hospital is 48.0 years old. suppose the distribution of ages is normal and has a standard deviation of 6.0 years. if 9 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 48.8 years. assume that the variable is normally distributed.

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Answers (1)

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Chaim Turner · Answer 1

True mean = mean (or average) + / - Z*SD/sqrt (sample population)

Now,

True mean has to be less than 48.8 years,

Mean (average) = 48.0 years,

SD = 6.0 years, and

Sample size (n) = 9 doctors

Making Z the subject of the formula

Z = (True mean - mean) / (SD/sqrt (n))

Substituting,

Z = (48.8-48.0) / (6.0/sqrt (9)) = 0.4

From Normal distribution probabilities table,

At Z = 0.4, P (x<0.4) = 0.6554 0r 65.54%