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Suppose that the weight, \(X\), in pounds, of a 40-year-old man is a normal random variable with a mean of 165 pounds and a standard deviation of 30 pounds. Determine \(P(X < 160)\). Round your answer to four decimal places.

Answer :

To find [tex]P(X < 160)[/tex] for a normal random variable [tex]X[/tex], which represents the weight of a 40-year-old man, we know the mean ([tex]\mu = 165[/tex] pounds) and the standard deviation ([tex]\sigma = 30[/tex] pounds). We will use the properties of the normal distribution to calculate this probability.

  1. Convert the raw score to a z-score:

    The z-score formula is:
    [tex]z = \frac{X - \mu}{\sigma}[/tex]

    Plug in the values:
    [tex]z = \frac{160 - 165}{30} = \frac{-5}{30} = -0.1667[/tex]

  2. Find the probability from the z-table:

A z-score of [tex]-0.1667[/tex] corresponds to a position on the standard normal distribution. Use a standard normal distribution table (z-table) to find [tex]P(Z < -0.1667)[/tex].

  • The z-table gives the probability that a standard normal random variable is less than the given z-score.

  • For [tex]z = -0.1667[/tex], the probability [tex]P(Z < -0.1667) \approx 0.4345[/tex].

  1. Interpret the result:

    • This means that there is approximately a 43.45% probability that a randomly selected 40-year-old man weighing according to this normal distribution weighs less than 160 pounds.

Therefore, the probability [tex]P(X < 160)[/tex] is approximately 0.4345 when rounded to four decimal places.

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