It is given that \(X \sim N(1.5, 3.2^2)\). Find the probability that a randomly chosen value of \(X\) is less than \(-2.4\).
Solution
To find the probability that \(X < -2.4\), we first standardize the variable using the formula for the standard normal distribution:
\(P(X < -2.4) = P\left( Z < \frac{-2.4 - 1.5}{3.2} \right)\)
Calculating the standardized value:
\(\frac{-2.4 - 1.5}{3.2} = \frac{-3.9}{3.2} = -1.219\)
Now, we find \(P(Z < -1.219)\) using the standard normal distribution table or a calculator:
\(P(Z < -1.219) = 1 - P(Z < 1.219)\)
From the standard normal distribution table, \(P(Z < 1.219) \approx 0.8886\).
Therefore, \(P(Z < -1.219) = 1 - 0.8886 = 0.1114\).
Rounding to three decimal places, the probability is \(0.111\).
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