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Question Video: Using Probabilities from Normal Distribution to Evaluate an Unknown Mathematics

Let 𝑋 be a random variable which is normally distributed with mean πœ‡ = 75 and standard deviation 𝜎 = 6. Given that 𝑃(𝑋 β‰₯ π‘˜) = 0.9938, find π‘˜.

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Video Transcript

Let 𝑋 be a random variable which is normally distributed with mean πœ‡ equals 75 and standard deviation 𝜎 equals six. Given that the probability 𝑋 is greater than or equal to π‘˜ equals 0.9938, find π‘˜.

We’re told that 𝑋 is a normal random variable with a mean of 75 and a standard deviation of six. We’re also given the probability that this normal random variable 𝑋 is greater than some unknown value π‘˜ is 0.9938. We can visualize this probability as the area to the right of the value of π‘˜ under the normal distribution curve. As this probability is greater than 0.5, we know that π‘˜ is in the lower half of the distribution, as the normal distribution is symmetrical about its mean. So the probability either side of the mean is 0.5.

We can deduce then that if the probability to the right of the mean is 0.5, then the probability that 𝑋 is greater than or equal to π‘˜ and less than or equal to the mean of 75 is 0.9938 minus 0.5, which is 0.4938. Now, we’ll need to use statistical tables later. But in order to find probabilities for the normal distribution, we need to first standardize the distribution. That is, we need to convert to a normal distribution with a mean of zero and a standard deviation of one.

We can convert any observation 𝑋 from any normal distribution with a mean of πœ‡ and a standard deviation 𝜎 to a standardized scale using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝜎. And these standardized values are known as 𝑧-scores. For the observation π‘˜ from a normal distribution with a mean of 75 and a standard deviation of six, the 𝑧-score is π‘˜ minus 75 over six. By standardizing, we find that the probability 𝑋 is greater or equal to π‘˜ and less than or equal to 75 is the same as the probability that the standard normal random variable 𝑍 is greater than or equal to π‘˜ minus 75 over six and less than or equal to zero. And we know this probability is 0.4938.

We’d like to use our statistical tables to now work out the value of π‘˜. But the tables give probabilities in a specific form. They give the probability that 𝑍 is between zero and a positive 𝑧-score. This is where we need to use the symmetry of the normal distribution. The normal distribution is symmetrical about its mean. So the probability that 𝑍 is between some negative value and zero is the same as the probability that 𝑍 is between zero and the absolute value of that negative value. So the probability that 𝑍 is between π‘˜ minus 75 over six and zero is the same as the probability that 𝑍 is between zero and negative π‘˜ minus 75 over six. This probability we know to be equal to 0.4938. So we can use our statistical tables.

Now, you may need to zoom in to see this. But if you do, you’ll see that a probability of 0.4938 is associated with a 𝑧-score of 2.5. This tells us then that negative π‘˜ minus 75 over six is equal to 2.5. And we can solve this equation to find the value of π‘˜. Multiplying both sides by negative one, we have π‘˜ minus 75 over six equals negative 2.5. We then multiply each side of the equation by six to give π‘˜ minus 75 equals negative 15. And finally we add 75 to each side to give π‘˜ is equal to 60.

This answer makes sense because 60 is less than 75. So the value of π‘˜ is in the lower half of the distribution. And we can be confident that the work we did in reflecting this region was correct. So by standardizing the normal distribution and then reflecting an area in the lower half of the distribution to become an area in the upper half of the distribution, we were able to use our statistical tables and then rearrange the equation this gave to find that the value of π‘˜ is 60.

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