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Question Video: Calculating Probability for a Normal Random Variable Mathematics

Let 𝑋 be a normal random variable. Find 𝑃(πœ‡ βˆ’ 0.56𝜎 < 𝑋 < πœ‡ + 1.64𝜎).

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Video Transcript

Let 𝑋 be a normal random variable. Find the probability that 𝑋 is greater than πœ‡ minus 0.56𝜎 and less than πœ‡ plus 1.64𝜎.

We’re told that 𝑋 is a normal random variable. And in general, we write the normal distribution as having a mean of πœ‡ and a standard deviation of 𝜎. Let’s begin by visualizing this probability we’re given as an area under the normal distribution curve. The normal distribution is symmetrical about its mean value πœ‡. The lower endpoint of this interval is πœ‡ minus 0.56𝜎. So, this value is 0.56 standard deviations below the mean; it’s in the lower half of the distribution. The upper endpoint is πœ‡ plus 1.64𝜎. So, this is 1.64 standard deviations above the mean; it’s in the upper half of the distribution.

We’re looking for the probability that 𝑋 is between these two values which corresponds to the area under the normal curve between these two limits. Now, in order to find this probability, we first need to convert these values to a standardized scale. We do this using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝜎. The value of 𝑧 we obtain is known as a 𝑧-score and is an observation from the standard normal distribution, which has mean zero and standard deviation one.

For the lower value πœ‡ minus 0.56𝜎, the 𝑧-score is πœ‡ minus 0.56𝜎 minus πœ‡ over 𝜎. That’s negative 0.56𝜎 over 𝜎, which is negative 0.56. And in fact, this makes sense. 𝑧-scores tell us how many standard deviations above or below the mean an observation is. And the value of πœ‡ minus 0.56𝜎 is 0.56 standard deviations below the mean. So, the 𝑧-score is negative 0.56. We don’t need to formally calculate the 𝑧-score for the upper value if we don’t want to. A value of πœ‡ plus 1.64𝜎 is 1.64 standard deviations above the mean. So, the 𝑧-score is 1.64. So, we find that the probability that the normal random variable 𝑋 is greater than πœ‡ minus 0.56𝜎 and less than πœ‡ plus 1.64𝜎 is equal to the probability that the standard normal random variable 𝑧 is greater than negative 0.56 and less than 1.64.

We can then divide this probability into the sum of two probabilities: the probability that 𝑧 is greater than negative 0.56 and less than zero, which is the area to the left of the mean, and the probability that 𝑧 is greater than zero and less than 1.64, which is the area to the right. We’re going to look these probabilities up in our statistical tables, which give probabilities of the form the probability that the standard normal random variable 𝑧 is greater than or equal to zero and less than or equal to some positive value 𝑧.

We don’t need to be concerned about whether we’re using weak or strict inequalities. Because 𝑧 is a continuous random variable, the probability that 𝑧 is actually equal to any particular value is zero. And so, the weak and strict inequality signs are interchangeable. Now, the probability corresponding to the area shaded in pink is in the correct format to use these statistical tables. But what about the probability corresponding to the area shaded in orange? Well, to work this out, we need to recall that the standard normal distribution is symmetrical about its mean. So, the probability that 𝑧 is between negative 0.56 and zero is the same as the probability that 𝑧 is between zero and positive 0.56. And this probability is in a form we can use to look up in our statistical tables.

Looking up the 𝑧-score of 1.64 in the tables first, we find the probability that 𝑧 is between zero and 1.64 is 0.4495. Next, looking at the probability associated with a 𝑧-score of 0.56, we find the probability that 𝑧 is between zero and 0.56 is 0.2123. And this is the same as the probability that 𝑧 is between negative 0.56 and zero. The total area is the sum of these two probabilities, which is 0.6618.

We’ve found then that the probability that the standard normal random variable 𝑧 is between negative 0.56 and 1.64, which is the same as the probability that the normal random variable 𝑋 is between πœ‡ minus 0.56𝜎 and πœ‡ plus 1.64𝜎 is 0.6618.

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