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]
