4.The amount of atmospheric radiation to which a person is exposed to while flying by jet across the country is considered a random variable with an approximate normal distribution. On a New York to Los Angeles flight, this distribution has an assumed mean of 5.2 mrem, with a standard deviation of 0.615. What is the probability that a randomly selected individual on such a flightis exposed to more than 6.0 mrem (millirem) of atmospheric radiation?

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ChiKesselman ChiKesselman · Answer 1 · 2020-03-09T05:14:19+01:00

Answer:

0.097 is the probability that a randomly selected individual is exposed to more than 6.0 mrem of atmospheric radiation.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5.2 mrem

Standard Deviation, σ = 0.615

We are given that the distribution of atmospheric radiation is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(flightis exposed to more than 6.0 mrem)

P(x > 6.0)

[tex]P( x > 6.0) = P( z > \displaystyle\frac{6.0 - 5.2}{0.615}) = P(z > 1.3008)[/tex]

[tex]= 1 - P(z \leq 1.3008)[/tex]

Calculation the value from standard normal z table, we have,

[tex]P(x > 6.0) = 1 - 0.903 =0.097= 9.7\%[/tex]

0.097 is the probability that a randomly selected individual is exposed to more than 6.0 mrem of atmospheric radiation.