Question

a) A type of battery-operated led lights has a known mean lifetime 7.8 hrs with standard deviation 0.5. It's provided that the lifetimes of these led lights are normally distributed. Without using the LSND program, find the probability that one of these led lights, selected at random, having lifetime between 6.8 and 7.8 hours.

b) According to a medical study, it takes about 48 hrs, on average, for Roseola virus to fade away in infected patients (children). Assuming that the fading times of Roseola virus in infected patients are normally distributed with standard deviation 0.7, what's the probability of having a random case where the fading time of the virus in an infected patient is less than 48 hours? Find the answer without using the LSND program.

c) pH measurements of a chemical solutions have mean 6.8 with standard deviation 0.04. Assuming all pH measurements of this solution follow a normal distribution, find the probability of selecting a pH measurement at random that reads below 6.68 OR above 6.88 without using the LSND program.

d) Salaries for senior Civil engineers per year have known mean of 72 K dollars and standard deviation 3.3 K dollars. Provided that the salaries per year for senior Civil engineers are normally distributed, what's the probability of finding a Civil engineer whose salary is between 68.7 K and 81.9 K dollars? Find the answer without using the LSND program.

e) Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are normally distributed, with a mean of 4.4 millimeters (mm) and a standard deviation of 1 mm. What's the probability that an ancient prehistoric Native American pot shard discovered in this area has thickness measurement thicker than 4.4? Find the answer without using the LSND program.

f)  scores in a Math class with large number of students have mean 147 and standard deviation 7.5. Provided the scores of this scores follow a normal distribution, what's the probability that a student scores below 154.5 OR above 162? Find the answer without using the LSND program.

Answer 1

A)

µ =    7.8                              
σ =    0.5                              
we need to calculate probability for ,                                  
P (   6.8   < X <   7.8   )                  
=P( (6.8-7.8)/0.5 < (X-µ)/σ < (7.8-7.8)/0.5 )                                  
                                  
P (    -2.000   < Z <    0.000   )                   
= P ( Z <    0.000   ) - P ( Z <   -2.000   ) =    0.5000   -    0.0228   =    0.4772

B)

µ =    48      
σ =    0.7      
          
P( X ≤    48   ) = P( (X-µ)/σ ≤ (48-48) /0.7)  
=P(Z ≤   0.000   ) =   0.5

C)

µ =    6.8                              
σ =    0.04                              
we need to calculate probability for ,                                  
P (   6.68   < X <   6.88   )                  
=P( (6.68-6.8)/0.04 < (X-µ)/σ < (6.88-6.8)/0.04 )                                  
                                  
P (    -3.000   < Z <    2.000   )                   
= P ( Z <    2.000   ) - P ( Z <   -3.000   ) =    0.9772   -    0.0013   =    0.9759

= 1- 0.9759

= 0.0241

D)

µ =    72                              
σ =    3.3                              
we need to calculate probability for ,                                  
P (   68.7   < X <   81.9   )                  
=P( (68.7-72)/3.3 < (X-µ)/σ < (81.9-72)/3.3 )                                  
                                  
P (    -1.000   < Z <    3.000   )                   
= P ( Z <    3.000   ) - P ( Z <   -1.000   ) =    0.9987   -    0.1587   =    0.84

e)

µ =    4.4              
σ =    1              
                  
P ( X ≥   4.40   ) = P( (X-µ)/σ ≥ (4.4-4.4) / 1)          
= P(Z ≥   0.000   ) = P( Z <   0.000   ) =    0.5

f)

µ =    147                              
σ =    7.5                              
we need to calculate probability for ,                                  
P (   154.5   < X <   162   )                  
=P( (154.5-147)/7.5 < (X-µ)/σ < (162-147)/7.5 )                                  
                                  
P (    1.000   < Z <    2.000   )                   
= P ( Z <    2.000   ) - P ( Z <   1.000   ) =    0.9772   -    0.8413   =    0.1359

=1- 0.1359

= 0.8641

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