Statistics Help?

A business wants to estimate the true mean annual income of its customers. The business needs to be within $250 of the true mean. The business estimates the true population standard deviation is around $2,400. If the confidence level is 90%, find the required sample size in order to meet the desired accuracy

The distribution of cholesterol levels in teenage boys is approximately normal with mean = 170 and standard deviation = 30 (Source: U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that a teenage boy has a cholesterol level greater than 200.

An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with mean = 15.5 and standard deviation = 3.6. What is the probability that during a given week the airline will lose between 10 and 20 suitcases?

A business wants to estimate the true mean annual income of its customers. The business needs to be within $250 of the true mean. The business estimates the true population standard deviation is around $2,400. If the confidence level is 90%, find the required sample size in order to meet the desired accuracy

m = margin of error = 250
standard deviation = 2400
90% confidence level
we need z*, at 90%, z* is 1.645

n = (1.645*2400/250)^2 = 249.39 and we will always round up to 250.
At least 250 samples.
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The distribution of cholesterol levels in teenage boys is approximately normal with mean = 170 and standard deviation = 30 (Source: U.S. National Center for Health Statistics). Levels above 200 warrant attention. Find the probability that a teenage boy has a cholesterol level greater than 200.

mean = 170
standard deviation = 30
x = 200
find z score, then seek for the corresponding probability value, p.
z = (200 - 170 )/ 30 = 30/30 = 1
with z = 1, p = 0.8413
you want to find the probability greater than 200, with p, we only have results under 200. Use the idea of under density curve, total area is 1.0
1.0 - 0.8413 = 0.1587 or 15.87%
0.1587 or 15.87% chance of being greater than 200 level.
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An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is approximately normal with mean = 15.5 and standard deviation = 3.6. What is the probability that during a given week the airline will lose between 10 and 20 suitcases?

mean = 15.5
standard deviation = 3.6
probability between 10 and 20?
use the same idea as above and find the individual probability then seek for the differences.
z at 10, z = (10 - 15.5 )/ 3.6= -1.53
z at 20, z = (20 - 15.5 )/ 3.6 = 1.53
with z at -1.53, p =0.0630
with z at 1.53, p = 0.9370

find the differences, 0.9370 - 0.0630 = 0.8740
or 87.40%
0.8740 will be you probability between 10 and 20 suitcases.
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