If the cosmic radiation to which a person is exposed while flying by jet across US is a random variable having mean 4.35 mrem and standard deviation 0.59 mrem,find the probabilities that the amount of cosmic radiation to which a person will be exposed on such a flight is:_______
(a) Between 4.00 and 5.00 mrem
(b) At least 5.50 mrem
(c) Less than 4.00 mrem

Respuesta :

Answer:

a

[tex]P(4.00 <  X  <  5.00) =  0.58818  [/tex]  

b

[tex]P(X \ge 5.5) = 0.02564 [/tex]

c

[tex]P(X < 4 ) = 0.27652[/tex]

Step-by-step explanation:

From the question we are told that

   The mean is  [tex]\mu = 4.35[/tex]

    The standard deviation is  [tex]\sigma =  0.59[/tex]

Generally the probability that the amount of cosmic radiation to which a person will be exposed on such a flight is between 4.00 and 5.00 mrem is mathematically represented as  

   [tex]P(4.00 <  X  <  5.00) =  P(\frac{ 4 - \mu }{\sigma} <  \frac{X - \mu}{\sigma} <  \frac{ 5 - \mu }{ \sigma}  )[/tex]

Here [tex]\frac{X - \mu}{\sigma}  = Z  (The \  standardized \  value \  of \  X )[/tex]

=> [tex]P(4.00 <  X  <  5.00) =  P(\frac{ 4 - 4.35 }{0.59} <  Z <  \frac{ 5 - 4.35 }{ 0.59}  )[/tex]

=> [tex]P(4.00 <  X  <  5.00) =  P(-0.59322 <  Z <  1.1017  )[/tex]

=> [tex]P(4.00 <  X  <  5.00) =  P( Z <  1.1017  ) - P(Z < -0.59322)  [/tex]

From the z -table the probability of  ( Z <  1.1017  ) and  (Z < -0.59322)   are

      [tex]P( Z <  1.1017  ) =0.8647[/tex]

and

      [tex]P( Z <  -0.59322 ) =0.27652[/tex]

So

=> [tex]P(4.00 <  X  <  5.00) =  0.8647 - 0.27652  [/tex]    

=>  [tex]P(4.00 <  X  <  5.00) =  0.58818  [/tex]    

Generally the probability that the amount of cosmic radiation to which a person will be exposed on such a flight is  At least 5.50 mrem is mathematically represented as

     [tex]P(X \ge 5.5) = 1- P(X < 5.5)[/tex]

Here

      [tex]P(X < 5.5) =  P(\frac{X - \mu }{\sigma}  <  \frac{5.5 - 4.35}{0.59}  )[/tex]

      [tex]P(X < 5.5) =  P(Z< 1.94915) [/tex]

From the z -table the probability of (Z< 1.94915) is  

     [tex]P(Z< 1.94915) = 0.97436[/tex]

So

       [tex]P(X \ge 5.5) = 1- 0.97436[/tex]

=>     [tex]P(X \ge 5.5) = 0.02564 [/tex]

Generally the probability that the amount of cosmic radiation to which a person will be exposed on such a flight is less than 4.00 mrem is mathematically represented as

    [tex]P(X < 4) =  P(\frac{X - \mu }{\sigma}  <  \frac{4 - 4.35}{0.59}  )[/tex]

      [tex]P(X < 4) =  P(Z< -0.59322) [/tex]

From the z -table the probability of (Z< 1.94915) is  

     [tex]P(Z< -0.59322) = 0.27652[/tex]

So

       [tex]P(X < 4 ) = 0.27652[/tex]