Let Z~N(0, 1). Find P(Z > -0.1).
Answer (2)
The probability that Z is greater than -0.1 is approximately 0.5398. This is found using the symmetry property of the normal distribution, which shows that P(Z > -0.1) is equal to P(Z < 0.1). Using the cumulative distribution function, we determined that the value of Φ(0.1) is about 0.5398.
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Recognize that -0.1)"> P ( Z > − 0.1 ) is the target probability for a standard normal distribution.
Apply the symmetry property: -0.1) = P(Z < 0.1)"> P ( Z > − 0.1 ) = P ( Z < 0.1 ) .
Use the CDF: P ( Z < 0.1 ) = Φ ( 0.1 ) .
Find the CDF value: -0.1) \approx \boxed{0.5398}"> P ( Z > − 0.1 ) ≈ 0.5398 .
Explanation
Understand the problem and provided data We are given that Z follows a standard normal distribution, denoted as Z ~ N(0, 1). Our goal is to find the probability that Z is greater than -0.1, which is written as P(Z > -0.1).
Use symmetry property To find P(Z > -0.1), we can use the properties of the standard normal distribution. One useful property is that the standard normal distribution is symmetric around 0. This means that the probability of Z being greater than -0.1 is the same as the probability of Z being less than 0.1. Mathematically, P(Z > -0.1) = P(Z < 0.1).
Express in terms of CDF We can express P(Z > -0.1) in terms of the cumulative distribution function (CDF) of the standard normal distribution, denoted by Φ(x). The CDF gives the probability that a standard normal random variable is less than or equal to x. Therefore, P(Z < 0.1) = Φ(0.1).
Find the CDF value The value of Φ(0.1) can be found using a standard normal distribution table or a calculator. The approximate value of Φ(0.1) is 0.5398. Therefore, P(Z > -0.1) = Φ(0.1) ≈ 0.5398.
State the final answer Therefore, the probability that Z is greater than -0.1 is approximately 0.5398.
Examples
Consider a scenario where you are analyzing the performance of students on a standardized test. If the test scores are normally distributed with a mean of 0 and a standard deviation of 1, then finding P(Z > -0.1) tells you the proportion of students who scored above -0.1 standard deviations from the mean. This can help you understand the distribution of scores and identify students who may need additional support. The probability is calculated as follows:
Since Z ∼ N ( 0 , 1 ) , we want to find -0.1)"> P ( Z > − 0.1 ) . Using the symmetry of the standard normal distribution, we know that -0.1) = P(Z < 0.1)"> P ( Z > − 0.1 ) = P ( Z < 0.1 ) .
Using a standard normal table or calculator, we find that P ( Z < 0.1 ) ≈ 0.5398 .
Therefore, -0.1) \approx 0.5398"> P ( Z > − 0.1 ) ≈ 0.5398 .
