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Suppose that the weight, [tex]$X$[/tex], in pounds, of a 40-year-old man is a normal random variable with a mean of 147 and a standard deviation of 16.

Determine [tex]$P(X\ \textless \ 185)$[/tex]. Round your answer to four decimal places.


Sagot :

To determine the probability [tex]\( P(X < 185) \)[/tex] for a normally distributed random variable [tex]\( X \)[/tex] with a mean ([tex]\(\mu\)[/tex]) of 147 pounds and a standard deviation ([tex]\(\sigma\)[/tex]) of 16 pounds, we follow these steps:

1. Find the z-score:
The z-score measures the number of standard deviations [tex]\( x \)[/tex] is from the mean. It can be calculated using the formula:

[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]

Where:
- [tex]\( x \)[/tex] is the value for which we are finding the probability ([tex]\( x = 185 \)[/tex] in this case).
- [tex]\( \mu \)[/tex] is the mean (147).
- [tex]\( \sigma \)[/tex] is the standard deviation (16).

Plugging in the values, we get:

[tex]\[ z = \frac{185 - 147}{16} = \frac{38}{16} = 2.375 \][/tex]

2. Consult the z-table or normal distribution calculator:
The next step is to determine the probability that a standard normal variable [tex]\( Z \)[/tex] is less than the calculated z-score ([tex]\( z = 2.375 \)[/tex]). This corresponds to finding [tex]\(P(Z < 2.375)\)[/tex].

3. Find the cumulative probability:
From standard normal distribution tables or using a statistical tool, the cumulative probability corresponding to [tex]\( z = 2.375 \)[/tex] is approximately 0.9912. This means there is a 99.12% probability that a randomly selected 40-year-old man weighs less than 185 pounds.

So, the probability [tex]\( P(X < 185) \)[/tex] is:

[tex]\[ P(X < 185) \approx 0.9912 \][/tex]

Thus, rounded to four decimal places, [tex]\( P(X < 185) = 0.9912 \)[/tex].