0 At a small metal fabrication company steel rods of a parti

0 At a small metal fabrication company, steel rods of a particular type cut to length have lengths with standard deviation 0.005 in. In an effort to determine the mean length produced using the current setup of the jig, a sample of rods is to be taken and their lengths measured, with the intention of using the value of X-bar as an estimate of µ. Approximate the probabilities that X-bar is within .0005 in. of µ for samples of size n = 50, 100 and 400

N= 50 P=?

N=100 P=?

N=400 P=?

Solution

The mean difference of the sampling distirbution to the mean is 0.

a)

N = 50:

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    -0.0005      
x2 = upper bound =    0.0005      
u = mean =    0      
n = sample size =    50      
s = standard deviation =    0.005      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -0.707106781      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    0.707106781      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.239750061      
P(z < z2) =    0.760249939      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.520499878   [answer]

************

b)

N = 100:

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    -0.0005      
x2 = upper bound =    0.0005      
u = mean =    0      
n = sample size =    100      
s = standard deviation =    0.005      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.158655254      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.682689492   [answer]

****************

c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    -0.0005      
x2 = upper bound =    0.0005      
u = mean =    0      
n = sample size =    400      
s = standard deviation =    0.005      
          
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   [answer]