Video: Estimating Population Percentages from a Normal Distribution in Context

The masses of a population of blackbirds are normally distributed with mean 103 g and standard deviation 11 g. To the nearest integer, what percentage of blackbirds have masses less than 110 g? To the nearest tenth, what percentage of blackbirds have masses greater than 124 g? To the nearest integer, what percentage of blackbirds have masses between 95 g and 120 g?

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Video Transcript

The masses of a population of blackbirds are normally distributed with mean 103 grams and standard deviation 11 grams. This is a context for a multipart question; the first part of which is β€œto the nearest integer, what percentage of blackbirds have masses less than 110 grams?”

Let’s call the distribution of the masses of the population of blackbirds 𝑋. We are told in the question that 𝑋 is normally distributed. So it follows a normal distribution with mean πœ‡ and variance 𝜎 squared. And we are further told that the mean is 103 grams and the standard deviation is 11 grams. So the variance is 11 squared.

So now that we’ve interpreted the context of the question, we can answer the first part. We’re asked to find the percentage of blackbirds that have a mass of less than 110 grams. Another way of asking that is to say what is the probability that if we draw a blackbird at a random from the population, we get a blackbird with a mass of less than 110 grams?

I’ve just written down this observation. Now, why is this observation useful? Well, because we have the distribution of the masses of the blackbirds, we know that this random variable 𝑋, which is the distribution of masses of the blackbirds, is normally distributed with mean 103 and standard deviation 11 or variance 11 squared. And so the proportion that we’re looking for is the probability that 𝑋 is less than 110.

How do you find this probability that 𝑋 is less than 110? Well, there are three possibilities, depending on your calculator. The first possibility is that the calculator you’re using gives a probability for any normal distribution. There is then a way of entering in the mean 103, the standard deviation 11, and the upper bound 110 into your calculator. And it just spits out the probability without you having to do any further work.

You may also be required to give the lower limit of the interval which you want to find the probability of. In this case, mathematically speaking, the lower limit is negative infinity. But in practice, we can replace this negative infinity by a really negative number, say negative 1000000, and we’ll get the right answer in practice. And if your calculator allows you to do this, then you should get an answer of 0.73773, something, something, something; it continues. And to the nearest integer percent, this is 74 percent.

Well, that’s great if your calculator can give the probability for any normal distribution, but not all calculators have this functionality. Some of them will only give the probability for the standard normal distribution. We still want to find the probability that the random variable 𝑋 is less than 110, but our calculator only allows us to find probabilities from the standard normal distribution, which we call 𝑍 distributed normally with mean zero and standard deviation one.

We need to turn our question about 𝑋 into a question about 𝑍. Well, it turns out that if 𝑋 is normally distributed with mean πœ‡ and variance 𝜎 squared, in other words with standard deviation 𝜎, then 𝑍 which is 𝑋 minus πœ‡, the mean, over 𝜎, the standard deviation, has the standard normal distribution that is the normal distribution with mean zero and standard deviation or variance one. And we can use this fact to change our question about the probability of 𝑋 being in some interval into a question about 𝑍 being in some interval, which of course we can then solve using our calculator.

The probability that 𝑋 is less than 110 is equal to the probability that 𝑋 minus 103 is less than 110 minus 103. This isn’t a fact about probability so much as it is a fact about inequalities. So these two inequalities have the same solution set and so they describe the same event. And so the probabilities of these two equal events is going to be the same.

Simplifying, we get this is the probability that 𝑋 minus 103 is less than seven. And dividing both sides of the inequality by 11, of course we don’t have to flip the inequality because 11 is positive, we get that this is equal to the probability that 𝑋 minus 103 over 11 is less than seven over 11.

Again, there’s no clever probability stuff going on; this is just about inequalities being the same or equivalent. And the reason it was useful to rewrite this inequality in this way is because 𝑋 was normally distributed with mean 103 and standard deviation 11. And we know there for that 𝑋 minus 103 over 11 is normally distributed with mean zero and standard deviation or variance one. So this is the probability that the standard normal distribution 𝑍 is less than seven over 11. And so because our calculator allows us to ask questions about the probability of a standard normal distribution 𝑍 being in some interval, we can use our calculator from here on.

In this case again, the lower limit about interval implicitly is negative infinity. And you might be required to choose some very small number to take the place of negative infinity when you put it into your calculator. Any number smaller than say negative four will do here. And of course, you should get the same answer as people whose calculator allowed them to skip this process, namely, 0.73773, and so on, which of course is 74 percent to the nearest integer percent.

If for whatever reason you can’t use your calculator for this question, the process is very similar to when the calculator only gives the probability for the standard normal distribution. We still need to find that the proportion we’re looking for is the probability that a standard normal distribution 𝑍 is less than seven over 11.

But now instead of using a calculator to find this probability, we have to use tables of values. And these tables of values hopefully for this part of the question normally give the probability that 𝑍 is less than some number. These tables don’t have values for 𝑍-scores like seven over 11. So we have to use a decimal approximation, in this case, 0.63. Better tables will have more decimal places in your 𝑍-score and therefore more accurate probabilities. But if we use a relatively basic table with only two decimal places allowed in the 𝑍-score, then we get a probability of 0.7357, which of course is still 74 percent to the nearest integer percent.

Now that we’ve solved the first part of the question, we can move on to the second part. To the nearest tenth, what percentage of blackbirds have masses greater than 124 grams?

We remember that 𝑋, the distribution of the masses of the blackbirds, is normally distributed with mean 103 and standard deviation 11. And the proportion of blackbirds with masses greater than 124 grams is the same as the probability that if you select a random blackbird from the population, its mass will be greater than 124 grams. So the number we’re looking for is the probability that 𝑋 is greater than 124. This is the probability that 𝑋 is between 124 and infinity.

And if we choose some very big number, let’s say 10 to the power of seven, to take the place of infinity, then if our calculator allows it, we can enter the lower and upper bounds for this interval along with the mean and standard deviation of our normal distribution. And we get that this probability is approximately 0.028125, and so on, which is 2.8 percent to the nearest tenth of a percent. So that’s our final answer to this part if we’re lucky enough to have a calculator with this functionality. If we are not so lucky, then we can use the fact that 𝑋 minus its mean, 103, divided by its standard deviation, 11, is normally distributed with mean zero and standard deviation one and hence variance one. This is the standard normal distribution which many calculators that don’t allow arbitrary normal distributions do allow.

To make use of this fact, we need to rewrite the inequality inside our probability in the form something is less than 𝑋 minus 103 over 11 is less than something else. So 𝑋 minus 103 over 11 is between two values, which we need to find. We can take the first bit of the inequality: 124 is less than 𝑋, subtract 103 from both sides to get 21 is less than 𝑋 minus 103, and then divide by 11. So we get 21 over 11 is less than 𝑋 minus 103 over 11. And we can do the same thing for the right-hand side: 𝑋 is less than infinity. Not much calculation is required here if 𝑋 is less than infinity. And all we can say is that 𝑋 minus 103 is less than infinity, and hence 𝑋 minus 103 over 11 is less than infinity.

So just replacing the inequality we had with this equivalent inequality, we get that the probability we’re looking for is the probability that 21 over 11 is less than 𝑋 minus 103 over 11 is less than infinity. And of course, its distribution of 𝑋 minus 103 over 11, which is conventionally called 𝑍, is normally distributed with mean zero and standard deviation one. This 21 over 11 is called the 𝑍-value. And entering it and some suitably large number in the place of infinity into your calculator, you should get the same answer as before; that is, 0.028125, and so on. And so the answer to the nearest tenth is 2.8 percent.

Alternatively, you can do this with a table of values for the standard normal distribution instead of a calculator. We want to find the probability that 𝑍 is greater than 21 over 11, but our tables only give the probability that 𝑍 is less than some value. So we use the fact that the probability that 𝑍 is greater than 21 over 11 is equal to one minus the probability that 𝑍 is less than 21 over 11. You might think that the less than sign here should be a less than or equal to sign strictly, but in fact it doesn’t matter because the probability that 𝑍 is equal to 21 over 11 is zero.

A normal distribution is a continuous distribution. And for any continuous distribution, the probability that the distribution takes one particular value is zero, there’re only positive probabilities on intervals of values. To use a table of values, we have to write our 𝑍-value as a decimal number with a certain number of decimal places depending on the accuracy of the table. So we take our 𝑍-value 21 over 11 and approximate it by 1.91. Looking this up in our table, we find the probability is one minus 0.9719, which is 0.0281. And so again, we get an answer of 2.8 percent to the nearest tenth of a percent.

And finally, to the nearest integer, what percentage of blackbirds have masses between 95 grams and 120 grams?

So 𝑋 is normally distributed with mean 103 and standard deviation 11. And we want to find the probability that 𝑋 is between 95 and 120. If we have a calculator allowing such a thing, we can just put in the mean standard deviation lower and upper bounds for intervals and get an answer of 0.70535, and so on, which written as a percentage to the nearest percent is 71 percent. If our calculator only allows us to find probabilities from the standard normal distribution with mean zero and standard deviation one, then we need to think about the distribution of 𝑋 minus 103 over 11, which is distributed with the standard normal distribution.

The region that we’re interested in is 95 is less than 𝑋 is less than 120. Taking just the first bit of this, the inequality 95 is less than 𝑋, and subtracting 103 from both sides, we get that negative eight is less than 𝑋 minus 103. And so negative eight over 11 is less than eight minus 103 over 11. Here, we’ve divided both sides by 11.

Focusing now on the right-hand side of this inequality, 𝑋 is less than 120, it’s the same procedure. So 𝑋 is less than 120, and so 𝑋 minus 103 is less than 17, and so 𝑋 minus 103 over 11 is less than 17 over 11. And so just replacing this inequality with the equivalent, we have the probability that negative eight over 11 is less than 𝑋 minus 103 over 11 is less than 17 over 11. And if we let 𝑍 be the distribution 𝑋 minus 103 over 11 as its convention, then we get the probability that negative eight over 11 is less than 𝑍 is less than 17 over 11. And with this standard normal distribution, we get the same answer of 0.70535, and so on, which is 71 percent to the nearest integer percent.

Alternatively, using tables of values, we have to rewrite our 𝑍-scores, negative eight over 11 and 17 over 11, as decimals with two decimal places. And this is equal to the probability of 𝑍 being less than 1.55 minus the probability that 𝑍 is less than negative 0.73. The easiest way to see why this is true is to use a graph of the bell curve of the normal distribution. The probability corresponds to the area of the shaded region between our two 𝑍-values: negative 0.73 and 1.55.

And this is equal to the area of the purple region, which is the probability of 𝑍 being less than 1.55 minus the area of the orange region, which is the probability that 𝑍 is less than negative 0.73. Looking up these two values in our tables, which we can do because the tables give the probability that 𝑍 is less than some number, we get 0.9394 minus 0.2327, which is 0.7067, which as a percentage to the nearest percent is like the other methods gave 71 percent.

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