a)Why is the distribution Normal and not right-skewed like the population? b)Why

Question

a)Why is the distribution Normal and not right-skewed like the population?

b)Why is the z-score 2?

c)What is the probability that the sample mean will be more than $3000 away from the population mean?

9.14 Income in Connecticut The average income in Connecticut in 2013 was $60,000 per person per year. tion is $30,000 Suppose the standard devia- and the distribution is right-skewed. Suppose we take a random sample of 400 residents of Connecticut. We want to find the probability that the sample mean will be more than $3000 away from the population mean. The TI-84 output is shown. NORMAL FLORT AUTO REAL RADIAN MP Z-Test 2-2 P .0455

Explanation / Answer

a)

This is because the central limit theorem states that the distribution of sample emans will be normal for sufficiently large enough sample size.

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b)

This is because the standard error is

SE = sigma/sqrt(n) = 30000/sqrt(400) = 1500

Hence,

z = (x-u)/SE = 3000/1500 = 2. [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 60000-3000=   57000      
x2 = upper bound = 60000+3000=   63000      
u = mean =    60000      
n = sample size =    400      
s = standard deviation =    30000      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.022750132      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.954499736      

Thus, those outside this interval is the complement =    0.045500264   [ANSWER]  

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