Answer:
p(z<0.42) = 0.6628
Step-by-step explanation:
A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. That is to say:
μ = 0
σ² = 1
Using a calculator, we find that:
p(z<0.42) = 0.6628 (See picture attached)
0.66276.
We are asked to find the approximate value of p(z<0.42) for a standard normal distribution.
Our given expression means the probability of getting a z-score less than 0.42.
We need to find the probability of getting the area corresponding to a z-score less than 0.42 under normal distribution curve.
We will normal distribution table to solve our given problem.
Therefore, the approximate value of our given expression would be 0.66276.
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Another way to name <UST is
<TSU
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