Let [tex]$X \sim N(-30.2,6)$[/tex]. Find [tex]$P(X > -32.9)$[/tex].
List the [tex]$z$[/tex]-scores needed to calculate the result. If there is more than one [tex]$z$[/tex]-score, separate the values with a comma.
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[tex]$P(X > -32.9)=[/tex]
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Answer (2)
To find -32.9)"> P ( X > − 32.9 ) for X ∼ N ( − 30.2 , 6 ) , we first calculated the z-score, which is − 1.102 . Then we used the standard normal distribution to find the probability, resulting in -32.9) \approx 0.865"> P ( X > − 32.9 ) ≈ 0.865 .
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Calculate the z-score: z = 6 − 32.9 − ( − 30.2 ) = − 1.102 .
Find the probability -1.102)"> P ( Z > − 1.102 ) using the standard normal distribution.
Use the property z) = 1 - P(Z < z)"> P ( Z > z ) = 1 − P ( Z < z ) .
Calculate the final probability: -32.9) = 1 - P(Z < -1.102) \approx \boxed{0.865}"> P ( X > − 32.9 ) = 1 − P ( Z < − 1.102 ) ≈ 0.865 .
Explanation
Analyze the problem and data We are given that X follows a normal distribution with mean μ = − 30.2 and variance σ 2 = 6 . We want to find the probability that X is greater than − 32.9 , i.e., -32.9)"> P ( X > − 32.9 ) . To do this, we will first calculate the z-score and then find the corresponding probability using the standard normal distribution.
Calculate the z-score The z-score is calculated using the formula: z = σ x − μ where x = − 32.9 , μ = − 30.2 , and σ = 6 .
Plugging in the values, we get: z = 6 − 32.9 − ( − 30.2 ) = 6 − 2.7 ≈ − 1.102
Find the probability using the z-score Now we need to find the probability -32.9)"> P ( X > − 32.9 ) , which is equivalent to finding -1.102)"> P ( Z > − 1.102 ) , where Z is a standard normal random variable. We can use the property that z) = 1 - P(Z < z)"> P ( Z > z ) = 1 − P ( Z < z ) .
So, we need to find 1 − P ( Z < − 1.102 ) . Using a standard normal distribution table or a calculator, we find that P ( Z < − 1.102 ) ≈ 0.135 . Therefore, -32.9) = 1 - 0.135 = 0.865"> P ( X > − 32.9 ) = 1 − 0.135 = 0.865
State the final answer Therefore, -32.9) \approx 0.865"> P ( X > − 32.9 ) ≈ 0.865 .
Examples
Consider a scenario where you are analyzing the performance of students on a test. The scores are normally distributed with a mean of -30.2 and a variance of 6. You want to find the probability that a student scores higher than -32.9. This calculation helps you understand how many students are likely to score above a certain threshold, which can be useful for evaluating the overall performance and identifying students who may need additional support.
