SOLUTION: The average age of statistics students nationwide is 22. The standard deviation is 2.5 years. Assume the age is a normally distributed variable.
Find the probability that on
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Question 313529: The average age of statistics students nationwide is 22. The standard deviation is 2.5 years. Assume the age is a normally distributed variable.
Find the probability that one student selected at random is older than 23.
Find the probability that the mean age of a group of 16 students selected at random is bigger than 23
can anyone please explain how to do this please
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The average age of statistics students nationwide is 22. The standard deviation is 2.5 years. Assume the age is a normally distributed variable.
Find the probability that one student selected at random is older than 23.
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Draw a normal curve with u = 22 ; let sigma = 2.5
Note: That illustrates the distribution of the ages of stat students.
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Now use the z-statistic to measure the probability above age 23.
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Convert 23 to a z-score:
z(23) = (23-22)/2.5 = 1/2.5 = 0.4
Note: This means that 23 is 0.4 standard deviations to the right of the mean.
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Now find the probability that a student is older than 23:
P(x> 23) = P(z > 0.4) = 0.3446
Note: You have to use your z-chart or some technology to find that number.
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Find the probability that the mean age of a group of 16 students selected at random is bigger than 23
Draw a normal curve with mean = 22 and std = 2.5/sqrt(16) = 0.625
Note: That is the distribution of the means of ALL samples of size 16.
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Find the z-score of 23 using u=22 and s=0.625
z(23) = (23-22)/0.625 = 1.6
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Find the probability a particular mean of a sample is greater than 23.
P(x-bar > 23) = P(z > 1.6) = 0.0548
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Cheers,
Stan H.
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