It is given that \(X \sim N(31.4, 3.6)\). Find the probability that a randomly chosen value of \(X\) is less than 29.4.
Solution
To find the probability that \(X < 29.4\), we first standardize the variable using the formula for the standard normal distribution:
\(P(X < 29.4) = P\left(Z < \frac{29.4 - 31.4}{\sqrt{3.6}}\right)\)
Calculating the standardized value:
\(\frac{29.4 - 31.4}{\sqrt{3.6}} = -1.0541\)
Thus, \(P(X < 29.4) = P(Z < -1.0541)\).
Using the standard normal distribution table, we find:
\(P(Z < -1.0541) = 1 - 0.8540 = 0.146\)
Therefore, the probability that a randomly chosen value of \(X\) is less than 29.4 is 0.146.
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