Question Video: Calculating Probability for an Interval of Finite Length for Standard Normal Random Variables Mathematics

Let 𝑍 be a standard normal random variable. Calculate 𝑃(βˆ’0.54 ≀ 𝑍 ≀ 2.33).

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Video Transcript

Let 𝑍 be a standard normal random variable. Calculate the probability that 𝑍 is greater than or equal to negative 0.54 and less than or equal to 2.33.

We’re told that 𝑍 is a standard normal random variable. So it has a mean of zero and a standard deviation of one. The interval for 𝑍 extends from a negative value to a positive value. We can visualize this probability as the area under the standard normal curve between the values of negative 0.54 and 2.33.

Remember, this curve is symmetrical about its mean, which for the standard normal is zero. So the value of negative 0.54 is in the lower half of the distribution. And the value of positive 2.33 is in the upper half. The area we want to calculate can be split into two: the area to the left of the mean, which is equivalent to the probability that 𝑍 is greater than or equal to negative 0.54 and less than or equal to zero, and the area to the right of the mean, which is equivalent to the probability that 𝑍 is greater than or equal to zero and less than or equal to 2.33.

We need to look up the probabilities corresponding to each of these areas in the standard normal distribution tables. But before we can do this, we need to ensure that the probabilities we’re looking for are in the correct format. There are various different types of table we can use. But the ones we’re going to use here give the probability that the random variable 𝑍 is between zero and a positive value 𝑍. This will be fine for the probability corresponding to the region shaded in pink. But what about the probability for the region shaded in orange?

We need to use the symmetry of the normal distribution curve to help with this. The probability that 𝑍 is between negative 0.54 and zero is the same as the probability that 𝑍 is between zero and positive 0.54. Essentially, we can reflect this area in the vertical axis of symmetry. And now this probability is in the correct format for us to use the standard normal tables.

Looking at the probability associated with a 𝑍-score of 2.33 first of all, we find that the probability 𝑍 is greater than or equal to zero and less than or equal to 2.33 is 0.4901. Next, looking up the probability associated with a 𝑍-score of 0.54, we find that the probability 𝑍 is greater than or equal to zero and less than or equal to 0.54, which is the same as the probability 𝑍 is greater than or equal to negative 0.54 and less than or equal to zero, is 0.2054. The probability that 𝑍 is between negative 0.54 and 2.33 then is the sum of these two values, which is 0.6955.

So using statistical tables and the symmetry of the standard normal distribution curve, we found that the probability 𝑍 is greater than or equal to negative 0.54 and less than or equal to 2.33 is 0.6955.

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