This is way over my head Can you show how to solve this on a

This is way over my head..... Can you show how to solve this on a TI-84?

Assume a sample of 40 observations are drawn from a population with mean 20 and variance 2. Compute the following. (a) Mean (Round your answer to the nearest integer.) and variance of bar X (Round your final answer to two decimal places (e.g. 98.76)) (b)P(bar X 20.0) Round your final answer to three decimal places (e.g. 98.765). (d)P(19.0

Solution

a)

The population mean is also the mean of the sampling distribution. Hence,

Mean = 20 [answer]

and

var(Xbar) = var/n = 2/40 = 0.05 [answer]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    20      
u = mean =    20      
n = sample size =    40      
s = standard deviation =    1.414213562      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   0   ) =    0.5 [ANSWER]

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c)

This is the complement of part B,

P(x>20) = 1 - P(x<=20) = 1 - 0.5 = 0.5 [answer]

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d)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    19      
x2 = upper bound =    21      
u = mean =    20      
n = sample size =    40      
s = standard deviation =    1.414213562      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -4.472135955      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    4.472135955      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    3.87211E-06      
P(z < z2) =    0.999996128      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.999992256 = 1.00 [answer]