Video: EG17S1-STATISTICS-Q09

EG17S1-STATISTICS-Q09

05:09

Video Transcript

If 𝑧 is a standard normally distributed variable where the probability that negative π‘˜ is less than or equal to 𝑧 is less than or equal to π‘˜ equals 0.8664, find π‘˜.

We’re told first of all that 𝑧 has a normal distribution. And a normal distribution is defined by its two parameters, its mean πœ‡ and its standard deviation 𝛿. We’re further told that 𝑧 has the standard normal distribution which means that its mean πœ‡ is equal to zero and its standard deviation 𝛿 is equal to one. So we can write that 𝑧 follows a normal distribution zero, one squared. We’re also told the probability that 𝑧 lies between two values, negative π‘˜ and π‘˜. And we want to work out the value of π‘˜. To do so, let’s recall how we find probabilities for a normal distribution.

We recall first of all that the normal distribution is a bell-shaped curve symmetrical about its mean πœ‡. The area below the full curve is equal to one. And the area to the left of a particular value gives the probability that the random variable 𝑧 takes a value less than or equal to this value. We can work out these probabilities by first calculating as 𝑧-score. That tells us how many standard deviations a particular value is away from the mean.

If we were working with a normal distribution other than the standard normal, we would calculate the 𝑧-score for any particular value π‘₯ using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝛿. We subtract the mean from the value π‘₯ and then divide by the standard deviation. However for the standard normal, the mean πœ‡ is zero. And the standard deviation 𝛿 is one. So we have π‘₯ minus zero over one which is just π‘₯. For a standard normal, the 𝑧-score of a particular value is just that value itself. Once we’ve calculated the 𝑧-score for a particular value, we can use our standard normal tables in order to look at the probability associated with that 𝑧-score.

However, we haven’t actually been given the probability that 𝑧 is less than or equal to a particular value. Instead, we’ve been given the probability that 𝑧 is between two values, negative π‘˜ and π‘˜. So how can we work backwards from knowing this probability to calculating the value of π‘˜? Well, there are actually a couple of different methods that we can use depending on the type of statistical tables that are available to us. For the first method, we’ll use tables which give the probability that the standard normally distributed variable 𝑧 is between zero and a value 𝑧. That would correspond to just the area now shaded in pink on our normal distribution curve. So we need to know the probability associated with just this area.

Well, to work this out, we recall again that our normal distribution is symmetrical about its mean, which means that the orange and pink areas are exactly the same. They’re each half of the probability of 0.8664. So they’re both equal to 0.4332. Now, here is that first type of statistical table. And we need to look up our probability of 0.4332. That probability is located here in our table. And if we look horizontally across, we see that it is associated with the 𝑧-score of 1.5. If we look vertically upwards, then we can see that the contribution in the second decimal place is zero. That tells us that the 𝑧-score which, in this case, is just a value of π‘˜ is 1.50 or 1.5.

So we have our answer π‘˜ equals 1.5. This tells us that the probability that 𝑧 is between 1.5 standard deviations below the mean and 1.5 standard deviations above the mean is 0.8664. But there is another method that we can use. This is to use tables which give the probability that 𝑧 is less than or equal to a particular value. On our normal distribution curve, this would mean we’re looking at all of the area to the left of that value π‘˜. So that’s the orange area and the pink area combined. We know that the pink area corresponds to a probability 0.4332. But what about our new orange area?

Well, we recall again that the normal distribution is symmetrical about its mean which means that the area on either side of the mean is half of the total area of one. The orange area is, therefore, equal to 0.5. This means that the total area to the left of π‘˜ and, therefore, the probability that 𝑧 is less than or equal to π‘˜ is 0.5 plus 0.4332 which is 0.9332. If we were to look up a probability of 0.9332 in our second type of statistical tables, we would again see that this is associated with the 𝑧-score of 1.5. We found then that if 𝑧 is a standard normally distributed variable where the probability that negative π‘˜ is less than or equal to 𝑧 is less than or equal to π‘˜ is 0.8664, then π‘˜ is equal to 1.5.

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