The average age of the residents in a city is 40 and the standard deviation is 17 years. The distribution of ages is known to be normal. Suppose a group of 15 people is formed to represent all age groups. The average age of this group is 50. What is the chance that the average age of a randomly selected group of 15 people from this population is at least 50 years old (round off to third decimal place)

P( \bar{x} >= 50) = 0.011

Step-by-step explanation:

From the question given, the details below were provided:

\mu = 40, \sigma = 17

Based on the central limit theorem,

P(\bar{x} < x) = P( Z < x - \mu / \sigma / sqrt(n) )

Thus,

P( \bar{x} >= 50) = P( Z >= 50 - 40 / 17 / sqrt(15) )

P( \bar{x} >= 50) = P( Z >= 2.2782)

P( \bar{x} >= 50) = 1 - P (Z < 2.2782)

P( \bar{x} >= 50) = 1 - 0.9886

P( \bar{x} >= 50) = 0.011

73.0769230769% of the games were won

y=mx+b

step-by-step explanation:

m slope, b y-intercept