Question

assume that the random variable x is normally distributed, with mean \\( \mu = 48 \\) and standard deviation \\( \sigma = 9 \\). compute the probability. be sure to draw a normal curve with the area corresponding to the probability shaded. \\( p(x \leq 46) \\) which of the following shaded regions corresponds to \\( p(x \leq 46) \\)?

Explanation:

Step1: Compute the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Given \(\mu = 48\), \(\sigma=9\), and \(x = 46\).

$$ z=\frac{46 - 48}{9}=\frac{-2}{9}\approx - 0.22 $$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\leq46)\), which is equivalent to \(P(Z\leq - 0.22)\) (after standardizing).
Looking up the value of \(z=-0.22\) in the standard - normal distribution table (the cumulative - distribution function for the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\leq - 0.22)=0.4129\)

Answer:

The probability \(P(X\leq46)\approx0.4129\). Regarding the normal - curve shading: The region to the left of \(x = 46\) (since \(P(X\leq46)\) represents the area under the normal curve from \(-\infty\) to \(x = 46\)) corresponds to option C (assuming the options are: A is a very narrow region around \(x = 46\) close to the mean, B is the region from the mean to \(x = 46\) on the right - hand side which is wrong as \(46<48\), and C is the region to the left of \(x = 46\)).