for a standard normal distribution, find the approximate value of $p(z\\leq0.42)$. use the portion of the…

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.\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.22 | 0.5871 |\n| 0.32 | 0.6255 |\n| 0.42 | 0.6628 |\n| 0.44 | 0.6700 |\n| 0.64 | 0.7389 |\n| 0.84 | 0.7995 |\n| 1.00 | 0.8413 |\n16%\n34%\n66%\n84%

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.\n| z | probability |\n| ---- | ---- |\n| 0.00 | 0.5000 |\n| 0.22 | 0.5871 |\n| 0.32 | 0.6255 |\n| 0.42 | 0.6628 |\n| 0.44 | 0.6700 |\n| 0.64 | 0.7389 |\n| 0.84 | 0.7995 |\n| 1.00 | 0.8413 |\n16%\n34%\n66%\n84%

Answer

Explanation:

Step1: Locate z - value in table

Find 0.42 in the z - column.

Step2: Read corresponding probability

The probability value for z = 0.42 is 0.6628.

Step3: Convert to percentage

0.6628×100% = 66.28% ≈ 66%

Answer:

66%