Suppose that 𝑋 is a normally distributed variable with the mean 𝜇 and a standard deviation 𝜎. Find probability that 𝑋 is greater than 𝜇 minus 1.5𝜎 and the probability that 𝜇 minus 1.96𝜎 is less than 𝑋 which is less than 𝜇 plus 1.96𝜎
Starting with the first one, the probability that 𝑋 is greater than 𝜇 minus 1.5𝜎, because we know that 𝑋 is a normally distributed variable, its probabilities are symmetrical about the mean under this bell-shaped curve. And that means the probability that 𝑋 will be less than the mean is 0.5, and the probability that 𝑋 will be greater than the mean is 0.5.
From there, we move away from the mean by counting the standard deviations, one standard deviation away from the mean, two standard deviations away from the mean, and on the left, a negative standard deviation, a negative two standard deviation. We want to know the probability that 𝑋 is greater than the mean minus 1.5 times a standard deviation. To do this, we’ll need to calculate a 𝑧-score.
To do this, we’ll need to calculate a 𝑧-score. To convert our 𝑋-value to a 𝑧-score, we take 𝑋 subtract the mean and then divide that by the standard deviation. In this case, we’ll need to take the mean 𝜇 minus 1.5𝜎 minus the mean. Since we’re adding 𝜇 and then subtracting 𝜇, those cancel out, leaving us with negative 1.5𝜎 over 𝜎. And those 𝜎s also cancel out, telling us we’re looking for the place where 𝑧 is greater than negative 1.5.
𝑧-scores also fall on this curve. And instead of using the mean and the standard deviation, we have a 𝑧-score of zero at the mean. And we’re interested in the places where 𝑧 is greater than negative 1.5, which is this probability. It’s the space under the curve up until the point negative 1.5. Before we move on, I want you to notice that when our 𝑋 is in the form 𝜇 plus or minus some standard deviation, the 𝑧-score is always the coefficient of 𝜎. This will help us a little bit later on.
We’ll need to consult a table of areas. Now, this table of areas is giving us the space between zero and 𝑧. This is why you’ll notice that there are no negative values for 𝑧. There is one important fact that we need to remember. This curve is symmetrical about the origin. And that means the distance from zero to 𝑧 is the same as the distance from negative 𝑧 to zero. We also remember that the probability that 𝑋 is above the mean, or the probability that 𝑧 is above zero, is half, that means the probability that 𝑧 is greater than zero plus the probability that 𝑧 falls between zero and one-half.
Our table is giving us the space between zero and one and a half, which we know will be the same amount as between zero and negative one and a half. We know we have one-half. And we look on the table at 1.5. And we see it has an area of 0.4332. We need to add 0.5 plus 0.4332. The probability that 𝑋 is greater than the mean minus 1.5𝜎 equals 0.9332. We need to follow a similar procedure to find part b).
𝑋 is normally distributed about the mean. And because the limits on 𝑋 are in this format, the mean plus or minus something times the standard deviation, the coefficient of the standard deviation is going to be our 𝑧-score. We need to consider the probability that negative 9.6 is less than 𝑧, which is less than 1.96. Our center is at zero. We’re interested in this area.
Because we know that this curve is symmetrical, we can take the area between zero and 1.96, that’s the probability that 𝑧 would fall between zero and 1.96, and we just multiply that by two. We’ll bring back our table of areas. For the probability that 𝑧 falls between zero and some 𝑧-score, this time we’re interested in 1.96. We go to the 1.9 row and then the 0.06 column. The probability that 𝑧 falls between zero and 1.96 is 0.4750.
So, we need to multiply 0.4750 times two. When we do that, we get 0.9500. And we can simplify that to say 0.95. The probability that 𝑋 falls between the mean minus 1.96 times a standard deviation and the mean plus 1.96 times the standard deviation equals 0.95.