Questions: Random Variables and Distributions Standard normal values: Advanced 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(-1.21 ≤ Z ≤ c)=0.8715 Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.

Random Variables and Distributions
Standard normal values: Advanced

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(-1.21 ≤ Z ≤ c)=0.8715

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

Transcript text: Random Variables and Distributions Standard normal values: Advanced 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. \[ P(-1.21 \leq Z \leq c)=0.8715 \] Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places. $\square$

Solution

Solution Steps

Step 1: Understanding the Problem

We need to find the value of $ c $ such that the probability $ P(-1.21 \leq Z \leq c) = 0.8715 $ for a standard normal distribution $ Z $. This can be expressed using the cumulative distribution function $ \Phi $ as: \[ P(-1.21 \leq Z \leq c) = \Phi(c) - \Phi(-1.21) = 0.8715 \]

Step 2: Calculate $ \Phi(-1.21) $

Using the standard normal distribution table or calculator, we find: \[ \Phi(-1.21) \approx 0.1131 \]

Step 3: Set Up the Equation

Substituting $ \Phi(-1.21) $ into the equation gives: \[ \Phi(c) - 0.1131 = 0.8715 \] This simplifies to: \[ \Phi(c) = 0.8715 + 0.1131 = 0.9846 \]

Step 4: Find $ c $ Corresponding to $ \Phi(c) = 0.9846 $

We need to find the value of $ c $ such that $ \Phi(c) = 0.9846 $. By checking values incrementally, we find: