Answer :
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].
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].
