Given a distribution with a mean of 40 and standard deviatio

Given a distribution with a mean of 40 and standard deviation of 13,sample size of 49, calculate the following probabilities.

P ( _{x >37 )=}

P ( _{x  < 43 )=}

P(43 < _{x  < 45 )=}

Solution

1.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    37      
u = mean =    40      
n = sample size =    49      
s = standard deviation =    13      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.615384615      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -1.615384615   ) =    0.946886285 [ans]

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2.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    43      
u = mean =    40      
n = sample size =    49      
s = standard deviation =    13      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.615384615      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   1.615384615   ) =    0.946886285 [ans]
          
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3.

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    43      
x2 = upper bound =    45      
u = mean =    40      
n = sample size =    49      
s = standard deviation =    13      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    1.615384615      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    2.692307692      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.946886285      
P(z < z2) =    0.996452028      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.049565743   [ans]