For a standard normal distribution, find the approximate value of [tex]P(-0.78 \leq z \leq 1.16)[/tex]. Use the portion of the standard normal table below to help answer the question.

\begin{tabular}{|c|c|}
\hline
z & Probability \\
\hline
0.00 & 0.5000 \\
\hline
0.16 & 0.5636 \\
\hline
0.22 & 0.5871 \\
\hline
0.78 & 0.7823 \\
\hline
1.00 & 0.8413 \\
\hline
1.16 & 0.8770 \\
\hline
1.78 & 0.9625 \\
\hline
2.00 & 0.9772 \\
\hline
\end{tabular}

A. 22%

B. 68%

C. 78%



Answer :

To find the probability [tex]\( P(-0.78 \leq z \leq 1.16) \)[/tex] for a standard normal distribution, follow these steps:

1. Understand the Problem:
We need to find the probability that a standard normal variable [tex]\( z \)[/tex] is between -0.78 and 1.16.

2. Locate the Probabilities from the Z-Table:
From the given table, find the cumulative probabilities for z-values of -0.78 and 1.16.

- For [tex]\( z = -0.78 \)[/tex]:
The table gives the cumulative probability [tex]\( P(Z \leq 0.78) = 0.7823 \)[/tex]. Since we are dealing with a negative z-value, we use the property of the standard normal distribution that the distribution is symmetric about the mean (which is zero). Therefore, the cumulative probability for [tex]\( z = -0.78 \)[/tex] is [tex]\( P(Z \leq -0.78) = 1 - P(Z \leq 0.78) \)[/tex].
Consequently:
[tex]\[ P(Z \leq -0.78) = 1 - 0.7823 = 0.2177 \][/tex]

- For [tex]\( z = 1.16 \)[/tex]:
The table directly provides the cumulative probability:
[tex]\[ P(Z \leq 1.16) = 0.8770 \][/tex]

3. Calculate the Probability Between the Two Z-Values:
The probability [tex]\( P(-0.78 \leq z \leq 1.16) \)[/tex] is the difference between the cumulative probability at [tex]\( z = 1.16 \)[/tex] and the cumulative probability at [tex]\( z = -0.78 \)[/tex]:
[tex]\[ P(-0.78 \leq z \leq 1.16) = P(Z \leq 1.16) - P(Z \leq -0.78) \][/tex]

Using the values from the previous step:
[tex]\[ P(-0.78 \leq z \leq 1.16) = 0.8770 - 0.2177 = 0.6593 \][/tex]

Therefore, the approximate value of [tex]\( P(-0.78 \leq z \leq 1.16) \)[/tex] is [tex]\( 0.6593 \)[/tex].