Question 1060679: Can someone show me how to solve these problem below using the Central Limit Theorem?
A specific study found that the average number of doctor visits per year for people over 55 is 8 with a standard deviation of 2. Assume that the variable is normally distributed.
1.Identify the population mean.
2.Identify the population standard deviation.
3.Suppose a random sample of 15 people over 55 is selected. What is the probability that the sample mean is above 9?
4.Suppose a random sample of 100 people over 55 is selected. What is the probability that the sample mean will be below 7?
A study found that citizens spend on average $1950 per year on groceries with a standard deviation of $100. Assume that the variable is normally distributed.
5.Identify the population mean.
6.Identify the population standard deviation.
7.Find the probability that a sample of 20 citizens will have a mean less than $1800.
8.Find the probability that a sample of 500 citizens will have a mean greater than $2000.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! When we have a population that is normally distributed, we make decision/s based on the sample size
:
1) population mean(u) is 8
2) population standard deviation(std.dev.) is 2
3) sample size(n) is 15 < 30 ( that is, the sample size is small per the Central Limit Theorem(CLT), therefore we will use the t-distribution.
Note that the sample mean is the same as the population mean per CLT
t-value = (9 - 8) / (2/square root(15) = 1.9365
degrees of freedom = 15 - 1 = 14
we use the table of t-values to get probability
Probability(Pr) ( X > 9 ) = 1 - Pr ( X < 9 )
Pr ( X > 9 ) = 1 - 0.9634 = 0.0366 approximately 0.04
4) n = 100 > 30, we use normal distribution tables and compute z-score
z-score = (7 - 8) / 2 = -0.50
Pr ( X < 7 ) = 0.3085 approximately 0.31
:
5) population mean(u) is 1950
6) population std.dev. is 100
7) n = 20 < 30, we use the t-statistic
t-value = (1800 - 1950) / (100/square root(20) = -6.7082
Note we use the absolute value of the t-value with the t-tables, then subtract from 1
degrees of freedom = 20 - 1 = 19
Pr ( X < 1800 ) = 1.00 - 1.00 = 0
8) n = 500 > 30, we use the z-score and the z-tables
z-value = ( 2000 - 1950 ) / 100 = 0.50
Pr ( X > 2000 ) = 1 - Pr ( X < 2000) = 1 - 0.6915 = 0.3085 approximately 0.31
|
|
|
