Question

Our bodies produce heat from the work of keeping us alive. This heat is known as the body temperature, which is continuous and approximately normally distributed. The previous standard for a healthy body temperature was 98.6 F. However, the current average body temperature for a healthy human is 98.2 F, with a population standard deviation of 0.75 F

Single Subject

1. What is the probability of selecting a person who has a healthy body temperature (between 97 F and 100 F)?
2. What is the probability of selecting someone who has a body temperature of at least 100 F?
3. Find and interpret

4. Take your body temperature (in degrees Fahrenheit) and state it:

   1. Use your body temperature from above to find your percentile for body temperature.
   2. Interpret the percentile value you found above.

Groups of Subjects

Now, assume that you are creating a sampling distribution of by sampling a group of people.

5. State the expected mean and standard deviation for body temperatures in a group of 5 people.
6. What is the probability of selecting a group of 5 people with an average body temperature below 97 F?
7. What is the probability of selecting a group of 5 people who have a body temperature of at least 98 F?
8. What is the probability of selecting a group of 15 people who have a body temperature of at least 98 F?
9. Describe what happens to the standard deviation of the sampling distribution as group size increases. Include mathematical justification.

please show work, i dont understand how to do this what so ever. tysm

Answer 1

1)

µ =    98.2                              
σ =    0.75                              
we need to calculate probability for ,                                  
97   ≤ X ≤    100                          
X1 =    97   ,   X2 =   100                  
                                  
Z1 =   (X1 - µ ) / σ =   -1.600                          
Z2 =   (X2 - µ ) / σ =   2.400                          
                                  
P (   97   < X <    100   ) =    P (    -1.6   < Z <    2.400   )
                                  
= P ( Z <    2.400   ) - P ( Z <   -1.600   ) =    0.9918   -    0.0548   =    0.9370
[excel formula for probability from z score =normsdist(z) ]

probability of selecting a person who has a healthy body temperature (between 97 F and 100 F)=0.9370

2)

µ =    98.2                          
σ =    0.75                          
right tailed                              
X ≥   100                          
                              
Z =   (X - µ ) / σ =   2.40                      
                              
P(X ≥   100   ) = P(Z ≥   2.40   ) =   P ( Z <   -2.40   ) =    0.0082
probability of selecting someone who has a body temperature of at least 100 F=0.0082

3)probability of selecting a person who has a healthy body temperature (between 97 F and 100 F)=0.9370

probability of selecting someone who has a body temperature of at least 100 F=0.0082

4)

leT my body temperature be 98.7 F

a)

µ =    98.2              
σ =    0.75              
left tailed                  
X ≤    98.7              
                  
Z =   (X - µ ) / σ =   0.67          
                  
P(X ≤   98.7   ) = P(Z ≤   0.67   ) =   0.7475

so, percentile = 74.75 th

b)

probability of selecting someone who has a body temperature fall below 98.7F=74.75%

5)

expected mean=µ=98.2F

expected std dev = σ/√n=0.75/√5=0.3354

6)

µ =    98.2              
σ =    0.75              
n=   5              
left tailed                  
X ≤    97              
                  
Z =   (X - µ )/(σ/√n) =   -3.578          
                  
P(X ≤   97   ) = P(Z ≤   -3.578   ) =   0.0002

probability of selecting a group of 5 people with an average body temperature below 97 F=0.0002

7)

µ =    98.2                          
σ =    0.75                          
n=   5                          
right tailed                              
X ≥   98                          
                              
Z =   (X - µ )/(σ/√n) =   -0.596                      
                              
P(X ≥   98   ) = P(Z ≥   -0.596   ) =   P ( Z <   0.596   ) =    0.7245

probability of selecting a group of 5 people who have a body temperature of at least 98 F=0.7245

8)

µ =    98.2                          
σ =    0.75                          
n=   15                          
right tailed                              
X ≥   98                          
                              
Z =   (X - µ )/(σ/√n) =   -1.033                      
                              
P(X ≥   98   ) = P(Z ≥   -1.033   ) =   P ( Z <   1.033   ) =    0.8492

probability of selecting a group of 15 people who have a body temperature of at least 98 F=0.8492

9)

std dev of sampling distribution =σ/√n

so, it is inversely proportional to sample size

so, when group size increases, std dev of sampling distribution decreases.

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excel formula for probability from z score =normsdist(z) ]