Let [tex]$X \sim N(-1.7,9.8)$[/tex]. Find [tex]$P(X\ \textless \ -15.2)$[/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.
[tex]$P(X\ \textless \ -15.2)= \square$[/tex]
Answer (2)
To find P ( X < − 15.2 ) where X ∼ N ( − 1.7 , 9.8 ) , we first calculate the standard deviation and then the z -score, which is approximately -4.3124. The probability P ( Z < − 4.3124 ) is approximately 0.00000807. Thus, P ( X < − 15.2 ) ≈ 0.00000807 .
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Calculate the standard deviation: σ = 9.8 ≈ 3.1305 .
Calculate the z -score: z = 9.8 − 15.2 − ( − 1.7 ) ≈ − 4.3124 .
Find the probability: P ( Z < − 4.3124 ) ≈ 0.00000807 .
The probability is approximately 0.00000807 and the z -score is − 4.3124 , so the final answer is 0.00000807 .
Explanation
Understand the problem and provided data We are given a normal distribution with mean μ = − 1.7 and variance σ 2 = 9.8 . We want to find the probability that X < − 15.2 . To do this, we first need to calculate the z -score.
Calculate the standard deviation The standard deviation is the square root of the variance, so σ = 9.8 ≈ 3.1305 .
Calculate the z-score The z -score is calculated as follows: z = σ X − μ = 9.8 − 15.2 − ( − 1.7 ) = 9.8 − 13.5 ≈ 3.1305 − 13.5 ≈ − 4.3124
Find the probability using the z-score Now we need to find the probability that Z < − 4.3124 , where Z is a standard normal random variable. This value is very small since − 4.3124 is far in the left tail of the standard normal distribution. Using a calculator or a standard normal distribution table, we find that P ( Z < − 4.3124 ) ≈ 0.00000807 .
State the final answer Therefore, P ( X < − 15.2 ) ≈ 0.00000807 . The z -score needed to calculate this result is approximately − 4.3124 .
Examples
In manufacturing, suppose you produce metal rods where the length is normally distributed. If you want to know the probability that a rod is shorter than a certain length, you would use a similar calculation. This helps in quality control to determine how often rods are outside acceptable length limits.
