Question

assume the random variable x is normally distributed with mean \\( \mu = 50 \\) and standard deviation \\( \sigma = 7 \\). compute the probability. be sure to draw a normal curve with the area corresponding to the probability shaded. \\( p(34<x<60) \\) which of the following normal curves corresponds to \\( p(34<x<60) \\)?

Explanation:

Step1: Convert values to z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 34\), \(z_1=\frac{34 - 50}{7}=\frac{-16}{7}\approx - 2.29\).
For \(x = 60\), \(z_2=\frac{60 - 50}{7}=\frac{10}{7}\approx1.43\).

Step2: Use the standard normal distribution table

We know that \(P(34\lt X\lt60)=P(-2.29\lt Z\lt1.43)\).
By the property \(P(a\lt Z\lt b)=\Phi(b)-\Phi(a)\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.
From the standard normal table, \(\Phi(-2.29) = 0.0110\) and \(\Phi(1.43)=0.9236\).
So \(P(-2.29\lt Z\lt1.43)=0.9236 - 0.0110=0.9126\).

Now, for the normal - curve part:
The normal curve has mean \(\mu = 50\). The interval \(34\lt X\lt60\) is symmetric around the mean in terms of the z - score calculation (but not symmetric in terms of raw values). The area between \(x = 34\) and \(x = 60\) should be shaded.
Looking at the options:

• Option A: Incorrect, because the shaded area is not between \(34\) and \(60\) (it has two non - adjacent shaded regions).
• Option B: Incorrect, because the un - shaded region is to the right of \(60\) (the area is not between \(34\) and \(60\)).
• Option C: Correct, because the shaded region is between \(34\) and \(60\).

Answer:

C.