I do not understand how to put the answer in the form of fra
I do not understand how to put the answer in the form of fraction.
A clinic took temperature readings of 250 flu patients over a weekend and discovered the temperature distribution to be Gaussian, with a mean of 101.70°F and a standard deviation of 0.5630. Use this normal error curve area table to find the following values. (a) What is the fraction of patients expected to have a fever greater than 103.39°F?
(b) What is the fraction of patients expected to have a temperature between 101.59°F and 102.21°F?
Ordinate and Area for a Normal Error Curve
 z = (x – )/
 The area refers to the area between z = 0 and z = the table value.
| |z| | y | Area | 
| 0.0 | 0.3989 | 0.0000 | 
| 0.1 | 0.3970 | 0.0398 | 
| 0.2 | 0.3910 | 0.0793 | 
| 0.3 | 0.3814 | 0.1179 | 
| 0.4 | 0.3683 | 0.1554 | 
| 0.5 | 0.3521 | 0.1915 | 
| 0.6 | 0.3332 | 0.2258 | 
| 0.7 | 0.3123 | 0.2580 | 
| 0.8 | 0.2897 | 0.2881 | 
| 0.9 | 0.2661 | 0.3159 | 
| 1.0 | 0.2420 | 0.3413 | 
| 1.1 | 0.2179 | 0.3643 | 
| 1.2 | 0.1942 | 0.3849 | 
| 1.3 | 0.1714 | 0.4032 | 
| 1.4 | 0.1497 | 0.4192 | 
| 1.5 | 0.1295 | 0.4332 | 
| 1.6 | 0.1109 | 0.4452 | 
| 1.7 | 0.0941 | 0.4554 | 
| 1.8 | 0.0790 | 0.4641 | 
| 1.9 | 0.0656 | 0.4713 | 
| 2.0 | 0.0540 | 0.4773 | 
| 2.1 | 0.0440 | 0.4821 | 
| 2.2 | 0.0355 | 0.4861 | 
| 2.3 | 0.0283 | 0.4893 | 
| 2.4 | 0.0224 | 0.4918 | 
| 2.5 | 0.0175 | 0.4938 | 
| 2.6 | 0.0136 | 0.4953 | 
| 2.7 | 0.0104 | 0.4965 | 
| 2.8 | 0.0079 | 0.4974 | 
| 2.9 | 0.0060 | 0.4981 | 
| 3.0 | 0.0044 | 0.498650 | 
| 3.1 | 0.0033 | 0.499032 | 
| 3.2 | 0.0024 | 0.499313 | 
| 3.3 | 0.0017 | 0.499517 | 
| 3.4 | 0.0012 | 0.499663 | 
| 3.5 | 0.0009 | 0.499767 | 
| 3.6 | 0.0006 | 0.499841 | 
| 3.7 | 0.0004 | 0.499904 | 
| 3.8 | 0.0003 | 0.499928 | 
| 3.9 | 0.0002 | 0.499952 | 
| 4.0 | 0.0001 | 0.499968 | 
| 0 | 0.5 | 
Solution
a)
We first get the z score for the critical value. As z = (x - u) / s, then as          
           
 x = critical value =    103.39      
 u = mean =    101.7      
           
 s = standard deviation =    0.563      
           
 Thus,          
           
 z = (x - u) / s =    3.001776199      
           
 Thus, using a table/technology, the right tailed area of this is          
           
 P(z >   3.001776199   ) =    0.001342047 [answer]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as          
 x1 = lower bound =    101.59      
 x2 = upper bound =    102.21      
 u = mean =    101.7      
           
 s = standard deviation =    0.563      
           
 Thus, the two z scores are          
           
 z1 = lower z score = (x1 - u)/s =    -0.195381883      
 z2 = upper z score = (x2 - u) / s =    0.905861456      
           
 Using table/technology, the left tailed areas between these z scores is          
           
           
 P(z < z2) =    0.817495406      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.817495406   [answer]  


