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
[tex]$z$[/tex] & 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. [tex]$22 \%$[/tex]

B. [tex]$66 \%$[/tex]

C. [tex]$78 \%$[/tex]



Answer :

Let's find the approximate value of [tex]\( P(-0.78 \leq z \leq 1.16) \)[/tex] using the provided portion of the standard normal table.

1. First, we need the cumulative probability corresponding to a [tex]\( z \)[/tex]-value of [tex]\( -0.78 \)[/tex]. However, the standard normal table provides the values for positive [tex]\( z \)[/tex]-values only. To find the probability for [tex]\( z = -0.78 \)[/tex], we use the symmetry of the normal distribution. For a negative [tex]\( z \)[/tex]-value, the cumulative probability is:

[tex]\[ P(Z \leq -0.78) = 1 - P(Z \leq 0.78) \][/tex]

From the provided table:
[tex]\[ P(Z \leq 0.78) = 0.7823 \][/tex]

Thus:
[tex]\[ P(Z \leq -0.78) = 1 - 0.7823 = 0.2177 \][/tex]

2. Next, we look up the cumulative probability for [tex]\( z = 1.16 \)[/tex] from the table:

[tex]\[ P(Z \leq 1.16) = 0.8770 \][/tex]

3. To find the probability that [tex]\( z \)[/tex] is between [tex]\( -0.78 \)[/tex] and [tex]\( 1.16 \)[/tex], we calculate:

[tex]\[ P(-0.78 \leq z \leq 1.16) = P(Z \leq 1.16) - P(Z \leq -0.78) \][/tex]

Substituting the values:

[tex]\[ P(-0.78 \leq z \leq 1.16) = 0.8770 - 0.2177 = 0.6593 \][/tex]

4. To express this probability as a percentage, we multiply by 100:

[tex]\[ P(-0.78 \leq z \leq 1.16) \times 100 = 65.93\% \][/tex]

Thus, the approximate value of [tex]\( P(-0.78 \leq z \leq 1.16) \)[/tex] is 65.93%. Out of the given options, the value closest to 65.93% is [tex]\( 66\% \)[/tex].