Video: MATH-STATS-2018-S1-Q3A

Consider a normal random variable 𝑋 with mean πœ‡ = 15 and standard deviation 𝜎 = 5. Find the value of 𝐾 such that 𝑃(𝑋 < 𝐾)= 0.1587.

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Video Transcript

Consider a normal random variable 𝑋 with mean πœ‡ equal to 15 and standard deviation 𝜎 equal to five. Find the value of 𝐾 such that the probability that 𝑋 is less than 𝐾 equals 0.1587.

A normal distribution is defined by two parameters, its mean, πœ‡, and its standard deviation, 𝜎. Both of which were given in this question. A normal distribution is a bell-shaped curve symmetrical about its mean. The area underneath the full curve is one. And the area to the left of a particular value represents the probability that 𝑋 is less than or equal to that value. However, as the normal distribution is a continuous distribution, this means that the probability that 𝑋 equals any particular value is effectively zero. So this probability is the same as the probability that 𝑋 is strictly less than that value.

Now we have statistical tables that can tell us these probabilities. But it’s impossible to have tables for every possible combination of values for the mean and standard deviation. Instead, we have one set of tables which tell us the probabilities for the standard normal distribution. This is the normal distribution with a mean of zero and a standard deviation of one. And we usually call this normally distributed random variable 𝑍. We can standardize values from our normal distribution by first subtracting the mean πœ‡ and then dividing by the standard deviation 𝜎. This will create a value taken from the standard normal distribution. And this value is known as the 𝑍-score. It tells us the number of standard deviations that a particular value π‘₯ is from the mean.

In our distribution, the mean, πœ‡, is 15. The standard deviation, 𝜎, is five. And the value of π‘₯ that we’re asked about is 𝐾. So we have 𝑍 equals 𝐾 minus 15 over five. We can then use an extract from our standard normal distribution tables to look up the 𝑍-score associated with the probability of 0.1587. In this extract from our standard normal distribution tables, we can see that the probability of 0.1587 is located here. And if we travel horizontally across, we see that it’s associated with a 𝑍-score of negative 1.00. Travelling vertically upwards, we can see that there are no hundredths to add on. So the 𝑍-score is exactly negative 1.00.

This means that the value 𝐾 for which the probability that 𝑋 is less than 𝐾 is 0.1587 is exactly one standard deviation below the mean. So we can substitute this value of negative 1.00, or just negative one, into our formula for the 𝑍-score. And it gives negative one equals 𝐾 minus 15 over five. To solve this equation for 𝐾, we must first multiply each side by five. And then, add 15 to both sides, giving 10 is equal to 𝐾. We know that this value is exactly one standard deviation below the mean as the mean was 15 and the standard deviation was five. And subtracting five from 15 gives 10. This means then that on the curve for our normal distribution, the area to the left of 10 is 0.1587. So we found the value of 𝐾. 𝐾 is equal to 10.

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