Let [tex]$Z \sim N(0,1)$[/tex]. Find [tex]$a$[/tex] so that [tex]$P(Z > a)=0.23$[/tex].
[tex]$a=$[/tex]
Answer (2)
To find a such that a) = 0.23"> P ( Z > a ) = 0.23 for a standard normal distribution, we first find the complementary probability: P ( Z ≤ a ) = 0.77 . Using the Z-table or inverse normal function, we find that a ≈ 0.7388 .
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Find the complementary probability: P ( Z ≤ a ) = 1 − 0.23 = 0.77 .
Use the inverse normal function (or Z-table) to find the z -score corresponding to a cumulative probability of 0.77.
The z -score is approximately a = 0.7388 .
Therefore, a = 0.7388 .
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 greater than a is 0.23, i.e., a) = 0.23"> P ( Z > a ) = 0.23 .
Find the complementary probability To find the value of a , we can use the standard normal distribution table (Z-table) or a calculator with the inverse normal function. The Z-table gives the cumulative probability P ( Z ≤ z ) for a given z -score. Since we have a) = 0.23"> P ( Z > a ) = 0.23 , we can find the complementary probability P ( Z ≤ a ) using the relationship: a)"> P ( Z ≤ a ) = 1 − P ( Z > a ) P ( Z ≤ a ) = 1 − 0.23 = 0.77
Find the z-score Now we need to find the value of a such that P ( Z ≤ a ) = 0.77 . We can use a Z-table or a calculator with the inverse normal function (also known as the quantile function) to find the corresponding z -score.
Determine the value of a Using a calculator with the inverse normal function, we find that the z -score corresponding to a cumulative probability of 0.77 is approximately 0.7388. Therefore, a ≈ 0.7388 .
Final Answer Thus, the value of a such that a) = 0.23"> P ( Z > a ) = 0.23 is approximately 0.7388.
Examples
Imagine you're designing a standardized test, and you want to set a cutoff score ( a ) such that only 23% of test-takers score above that level. Knowing that test scores follow a standard normal distribution, you can use this calculation to find the appropriate cutoff score. This ensures that the test is appropriately challenging and that the desired percentage of test-takers are identified as high-achievers. This is a practical application of understanding probabilities related to the standard normal distribution.
