Question Video: Using Probability for Standard Normal Random Variables to Find Unknown Variables | Nagwa Question Video: Using Probability for Standard Normal Random Variables to Find Unknown Variables | Nagwa

Question Video: Using Probability for Standard Normal Random Variables to Find Unknown Variables Mathematics

Suppose 𝑍 is a standard normal random variable. Given 𝑃(𝑍 ≀ π‘˜) = 0.9922, find the value of π‘˜.

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Video Transcript

Suppose 𝑍 is a standard normal random variable. Given the probability 𝑍 is less than or equal to π‘˜ equals 0.9922, find the value of π‘˜.

We’re told that 𝑍 is a standard normal random variable, so it has a normal distribution with a mean of zero and a standard deviation of one. We’re given that the probability that 𝑍 is less than equal to some unknown value π‘˜ is 0.9922. We can visualize this probability as the area to the left of the value of π‘˜ under the standard normal curve. As this probability is greater than 0.5, we know that π‘˜ is positive, as the area either side of the mean under the standard normal curve is equal to 0.5.

Using the symmetry of the normal distribution, we can separate this probability of 0.9922 into two probabilities: a probability of 0.5, which is represented by the area to the left of the mean of zero, and a probability of 0.4922, that’s 0.9922 minus 0.5, which is represented by the area to the right of the mean between zero and π‘˜. In other words, the probability that 𝑍 is greater than or equal to zero and less than or equal to π‘˜ is 0.4922.

And this is great because this probability is in the correct format for us to use our statistical tables. The type of tables we’re using here give the probability that the standard normal random variable 𝑍 is between zero and some positive value lowercase 𝑧. And you may also have a calculator which has the functionality to give you these values. We’re essentially using these tables in reverse. We find the value 0.4922 in the table. And we then see that it is associated with a 𝑍-score of 2.42.

So, by separating this probability into a probability for the area to the left of the mean and a probability for the area to the right of the mean, we found that the value of π‘˜ is 2.42.

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