Let Z~N(0, 1). Find a so that P(Z < a) = 0.69.
Answer (2)
To find 'a' such that P(Z < a) = 0.69 in a standard normal distribution (Z ~ N(0, 1)), we use the inverse CDF. The calculated value is approximately 0.4959. This indicates that about 69% of standard normal values lie below this point.
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We have a standard normal distribution Z ~ N(0, 1) and want to find 'a' such that P(Z < a) = 0.69.
Use the inverse CDF (quantile function) to find the value of 'a'.
Calculate a = Φ^{-1}(0.69) ≈ 0.4959.
The value of a is approximately 0.4959 .
Explanation
Understand the problem and provided data We are given that Z follows a standard normal distribution, denoted as Z ~ N(0, 1). This means Z has a mean of 0 and a standard deviation of 1. We want to find the value 'a' such that the probability of Z being less than 'a' is 0.69, i.e., P(Z < a) = 0.69.
Using the inverse CDF To find the value of 'a', we need to use the inverse of the standard normal cumulative distribution function (CDF), also known as the quantile function or probit function. This function tells us, for a given probability, what the corresponding value is in the standard normal distribution. In other words, we want to find the value 'a' such that a = Φ^{-1}(0.69), where Φ^{-1} is the inverse CDF of the standard normal distribution.
Find the value of a Using a calculator or a standard normal table, we find the value of 'a' that corresponds to a probability of 0.69. The result of this operation is approximately 0.4959.
Final Answer Therefore, the value of 'a' such that P(Z < a) = 0.69 is approximately 0.4959.
Examples
Imagine you're analyzing test scores that follow a normal distribution. If you want to find the score below which 69% of the students fall, you would use the same method. This is useful for setting grade boundaries or understanding the distribution of scores. For example, if the scores are normally distributed with a mean of 0 and a standard deviation of 1, then the score 'a' below which 69% of students fall is approximately 0.4959.
