Question

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for a standard normal distribution, find the approximate value of $p(z\leq0.42)$. use the portion of the standard normal table below to help answer the question.

z | probability
0.00 | 0.5000
0.22 | 0.5871
0.32 | 0.6255
0.42 | 0.6628
0.44 | 0.6700
0.64 | 0.7389
0.84 | 0.7995
1.00 | 0.8413

16%
34%
66%
84%
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Explanation:

Step1: Locate z=0.42 in the table

Find the row where $z=0.42$ in the provided standard normal table.

Step2: Match to probability value

The corresponding probability for $z=0.42$ is $0.6628$.

Step3: Convert to percentage

Multiply the decimal by 100: $0.6628 \times 100 = 66.28\%$, which approximates to 66%.

Answer:

C. 66%