Question

for a standard normal distribution, find the approximate value of ( p(z leq 0.42) ). use the portion of the standard normal table below to help answer the question.

( z )	probability
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%

Explanation:

Step1: Understand the standard normal distribution

In a standard normal distribution, \( F(z \leq a) \) represents the cumulative probability that a standard normal variable \( z \) is less than or equal to \( a \). This is given by the probability value corresponding to \( z = a \) in the standard normal table.

Step2: Find the probability for \( z = 0.42 \)

Looking at the provided standard normal table, when \( z = 0.42 \), the corresponding probability is \( 0.6628 \). To convert this to a percentage, we multiply by \( 100 \): \( 0.6628\times100 = 66.28\% \), which is approximately \( 66\% \).

Answer:

66% (corresponding to the option with 66%)