Questions: Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true. P(-c ≤ Z ≤ c)=0.9199 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.

Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true.
P(-c ≤ Z ≤ c)=0.9199

Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.

Transcript text: Suppose $Z$ follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of $c$ so that the following is true. \[ P(-c \leq Z \leq c)=0.9199 \] Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.

Solution

Solution Steps

Step 1: Understanding the Problem

We need to find the value of $ c $ such that the probability $ P(-c \leq Z \leq c) = 0.9199 $ for a standard normal distribution $ Z $. This can be expressed in terms of the cumulative distribution function $ \Phi $ as: \[ P(-c \leq Z \leq c) = \Phi(c) - \Phi(-c) \] Since the standard normal distribution is symmetric, we have $ \Phi(-c) = 1 - \Phi(c) $. Therefore, we can rewrite the equation as: \[ P(-c \leq Z \leq c) = \Phi(c) - (1 - \Phi(c)) = 2\Phi(c) - 1 \] Setting this equal to 0.9199 gives: \[ 2\Phi(c) - 1 = 0.9199 \]

Step 2: Solving for $ \Phi(c) $

Rearranging the equation, we find: \[ 2\Phi(c) = 0.9199 + 1 = 1.9199 \] \[ \Phi(c) = \frac{1.9199}{2} = 0.95995 \]

Step 3: Finding $ c $

Next, we need to find the value of $ c $ such that $ \Phi(c) = 0.95995 $. We can calculate $ \Phi(c) $ for various values of $ c $ and find the one that is closest to 0.95995. The output from the calculations shows the following probabilities for different $ c $ values:

$ \Phi(1.0) - \Phi(-1.0) = 0.6827 $
$ \Phi(1.1) - \Phi(-1.1) = 0.7287 $
$ \Phi(1.2) - \Phi(-1.2) = 0.7699 $
$ \Phi(1.3) - \Phi(-1.3) = 0.8064 $
$ \Phi(1.4) - \Phi(-1.4) = 0.8385 $
$ \Phi(1.5) - \Phi(-1.5) = 0.8664 $
$ \Phi(1.6) - \Phi(-1.6) = 0.8910 $
$ \Phi(1.7) - \Phi(-1.7) = 0.9115 $
$ \Phi(1.8) - \Phi(-1.8) = 0.9282 $
$ \Phi(1.9) - \Phi(-1.9) = 0.9418 $
$ \Phi(2.0) - \Phi(-2.0) = 0.9545 $
$ \Phi(2.1) - \Phi(-2.1) = 0.9644 $

From the calculations, we can see that $ \Phi(2.0) $ is approximately 0.9545, and $ \Phi(2.1) $ is approximately 0.9644. Therefore, the value of $ c $ that satisfies $ \Phi(c) \approx 0.95995 $ is between 2.0 and 2.1.

Step 4: Rounding $ c $

To find $ c $ to two decimal places, we can take the average of 2.0 and 2.1, which gives us: \[ c \approx 2.0 \] Rounding to two decimal places, we find: \[ c = 2.00 \]