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.
\[
P(-c \leq Z \leq c)=0.9676
\]
Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.
Answer
Correcting the calculation, we find that the value of $c$ that makes $P(-c \leq Z \leq c)=0.9676$ true is approximately \(\boxed{2.14}\).
Step 1 ：Suppose $Z$ follows the standard normal distribution. We need to determine the value of $c$ so that $P(-c \leq Z \leq c)=0.9676$.
Step 2 ：Since the standard normal distribution is symmetric about the mean, which is 0 for the standard normal distribution, the probability $P(-c \leq Z \leq c)$ is the same as the probability $P(0 \leq Z \leq c)$ multiplied by 2. So, we need to find the value of $c$ such that $2P(0 \leq Z \leq c) = 0.9676$.
Step 3 ：We can use the Z-table to find the Z-score that corresponds to a cumulative probability of $0.9676/2 = 0.4838$.
Step 4 ：However, the value of $c$ is negative, which is not possible in this context because we are looking for a positive value of $c$. This is because the Z-score is a measure of how many standard deviations an element is from the mean, and it cannot be negative in this context. The negative value indicates that there is an error in the calculation.
Step 5 ：Upon reviewing the calculation, we realize that we made a mistake in the calculation of the cumulative probability. The cumulative probability should be $0.9676/2 + 0.5 = 0.9838$, not $0.9676/2 = 0.4838$. This is because the cumulative probability for the standard normal distribution at Z=0 is 0.5, and we are looking for the Z-score that corresponds to a cumulative probability of 0.9838.
Step 6 ：Correcting the calculation, we find that the value of $c$ that makes $P(-c \leq Z \leq c)=0.9676$ true is approximately \(\boxed{2.14}\).
