Question

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

Answer 1

a

`P(4.00 <  X  <  5.00) =  0.58818`

b

`P(X \ge 5.5) = 0.02564`

c

`P(X < 4 ) = 0.27652`

Explanation:

From the question we are told that

The mean is
`\mu = 4.35`

The standard deviation is
`\sigma =  0.59`

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

`P(4.00 <  X  <  5.00) =  P(( 4 - \mu )/(\sigma) <  (X - \mu)/(\sigma) <  ( 5 - \mu )/( \sigma)  )`

Here
`(X - \mu)/(\sigma)  = Z  (The \  standardized \  value \  of \  X )`

=>
`P(4.00 <  X  <  5.00) =  P(( 4 - 4.35 )/(0.59) <  Z <  ( 5 - 4.35 )/( 0.59)  )`

=>
`P(4.00 <  X  <  5.00) =  P(-0.59322 <  Z <  1.1017  )`

=>
`P(4.00 <  X  <  5.00) =  P( Z <  1.1017  ) - P(Z < -0.59322)`

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

`P( Z <  1.1017  ) =0.8647`

and

`P( Z <  -0.59322 ) =0.27652`

So

=>
`P(4.00 <  X  <  5.00) =  0.8647 - 0.27652`

=>
`P(4.00 <  X  <  5.00) =  0.58818`

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

`P(X \ge 5.5) = 1- P(X < 5.5)`

Here

`P(X < 5.5) =  P((X - \mu )/(\sigma)  <  (5.5 - 4.35)/(0.59)  )`

`P(X < 5.5) =  P(Z< 1.94915)`

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

`P(Z< 1.94915) = 0.97436`

So

`P(X \ge 5.5) = 1- 0.97436`

=>
`P(X \ge 5.5) = 0.02564`

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

`P(X < 4) =  P((X - \mu )/(\sigma)  <  (4 - 4.35)/(0.59)  )`

`P(X < 4) =  P(Z< -0.59322)`

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

`P(Z< -0.59322) = 0.27652`

So

`P(X < 4 ) = 0.27652`