5.4 – Pulse Rates of Women. Women have pulse rates that are normally distributed

Question

5.4 – Pulse Rates of Women. Women have pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute (based on the data in the Lab 5 tables).

b) Dr. Puretz sees exactly 25 female patients each day. Find the probability that 25 randomly selected women have a mean pulse rate between 70 beats per minute and 85 beats per minute.

c) If Dr. Puretz wants to select pulse rates to be used as cutoff values for determining when further tests should be required, which pulse rates are better to use: the results from part (a) or the pulse rates of 70 beats per minute and 85 beats per minute from part (b)? Why?

For this last part read the question carefully & explain your answer.

Explanation / Answer

5.4.

a)

For P1:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.01      
          
Then, using table or technology,          
          
z =    -2.326347874      
          
As x = u + z * s,          
          
where          
          
u = mean =    77.5      
z = the critical z score =    -2.326347874      
s = standard deviation =    11.6      
          
Then          
          
x = P1 = critical value =    50.51436466   [ANSWER]

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For P99:

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.99      
          
Then, using table or technology,          
          
z =    2.326347874      
          
As x = u + z * s,          
          
where          
          
u = mean =    77.5      
z = the critical z score =    2.326347874      
s = standard deviation =    11.6      
          
Then          
          
x = P99 = critical value =    104.4856353   [ANSWER]  
  
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b)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    70      
x2 = upper bound =    85      
u = mean =    77.5      
n = sample size =    25      
s = standard deviation =    11.6      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -3.232758621      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3.232758621      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.000613006      
P(z < z2) =    0.999386994      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.998773989   [ANSWER]

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c)
  
Part a) is better, because part b) deals with 25 diffrent patients. However, here, we are dealing with a specific patient.

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