Jims systolic blood pressure is a random variable with a mea

Jim?s systolic blood pressure is a random variable with a mean of 146 mmHg and a standard deviation of 19 mmHg. For Jim?s age group, 140 is the threshold for high blood pressure. Assume the data given follows the normal probability distribution. (a) If Jim?s systolic blood pressure is taken at a randomly chosen moment, what is the probability that it will be 142 or less? (Round the value of z to 2 decimals. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimals.) Probability (b) If Jim?s systolic blood pressure is taken at a randomly chosen moment, what is the probability that it will be 179 or more? (Round the value of z to 2 decimals. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimals.) Probability (c) If Jim?s systolic blood pressure is taken at a randomly chosen moment, what is the probability that it will be between 121 and 169? (Round the value of z to 2 decimals. Use Appendix C-2 to find probabilities. Round your final answer to 4 decimals.) Probability

Solution

A.
          
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    142      
u = mean =    146      
n = sample size =    1      
s = standard deviation =    19      
          
Thus,          
          
z =    -0.21      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -0.21   ) =    0.4168 [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 =    179      
u = mean =    146      
n = sample size =    1      
s = standard deviation =    19      
          
Thus,          
          
z =    1.74      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.74   ) =    0.0409 [answer]

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We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    121      
x2 = upper bound =    169      
u = mean =    146      
n = sample size =    1      
s = standard deviation =    19      
          
          
          
z1 = lower z score =    -1.32      
z2 = upper z score =    1.21      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.0934      
P(z < z2) =    0.8869      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.7935   [answer]

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