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For a standard normal distribution, find the approximate value of [tex]\( P(z \leq 0.42) \)[/tex].

Use the portion of the standard normal table below:

[tex]\[
\begin{tabular}{|c|c|}
\hline
$z$ & Probability \\
\hline
0.00 & 0.5000 \\
\hline
0.22 & 0.5871 \\
\hline
0.32 & 0.6255 \\
\hline
0.42 & 0.6628 \\
\hline
0.44 & 0.6700 \\
\hline
0.64 & 0.7389 \\
\hline
0.84 & 0.7995 \\
\hline
1.00 & 0.8413 \\
\hline
\end{tabular}
\][/tex]

A. 16%

B. 34%

C. 66%

D. 84%

Sagot :

GinnyAnsweridn
Sure, let's find the approximate value of [tex]\(P(z \leq 0.42)\)[/tex] using the provided portion of the standard normal table.

1. Understand the Problem: We are asked to find the probability that a standard normally distributed variable [tex]\(z\)[/tex] is less than or equal to 0.42.

2. Table Interpretation: The standard normal table (or Z-table) provides the area (or probability) to the left of a given [tex]\(z\)[/tex]-value in a standard normal distribution, which has a mean of 0 and a standard deviation of 1.

3. Locate the [tex]\(z\)[/tex]-value: From the table provided, locate the row where [tex]\(z = 0.42\)[/tex].

4. Read the Probability: Corresponding to [tex]\(z = 0.42\)[/tex], the table entry is 0.6628.

Therefore, the approximate value of [tex]\(P(z \leq 0.42)\)[/tex] is 0.6628.

So, the correct choice here, in percentage form, would be [tex]\(66 \%\)[/tex].
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