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Let \(x\) be a continuous random variable that is normally distributed with a mean of 80 and a standard deviation of \(12 .\) Find the probability that \(x\) assumes a value a. greater than 69 b. less than 73 c. greater than 101 d. less than 87

Short Answer

Expert verified
The probabilities for a value greater than 69, less than 73, greater than 101, and less than 87 are approximately 0.8212, 0.2810, 0.0401, and 0.7190 respectively.

Step by step solution

01

Calculation of the Z-score for each part

We need to convert the given value into Z-scores using the formula \(Z=(X-μ)/σ\). Here, μ is the mean and σ is the standard deviation. For each of the four parts, we calculate the z-score:\na. \(Z = (69-80)/12 \approx -0.92\)\nb. \(Z = (73-80)/12 \approx -0.58\)\nc. \(Z = (101-80)/12 \approx 1.75\)\nd. \(Z = (87-80)/12 \approx 0.58\)
02

Lookup of the Z-scores in the Standard Normal Distribution Table

Once we have calculated the Z-score we can use the Z-table to find the probabilities. The Z-table shows the cumulative probabilities for the standard normal distribution.\na. For Z = -0.92, the table gives us the probability 0.1788, which is the cumulative probability from the left up to Z. But since we want to find the probability greater than 69, we take the complement: 1 - 0.1788 = 0.8212\nb. For Z = -0.58, the table gives us the cumulative probability 0.2810. Since we want less than 73, this is the probability we take.\nc. For Z = 1.75, the table gives us 0.9599. We want more than 101, so we take the complement: 1 - 0.9599 = 0.0401\nd. For Z = 0.58, the table gives us 0.7190. Since we want less than 87, this is the probability we take.

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