Video Transcript
Given that 𝑍 is a standard normal random variable and the probability that 𝑍 is greater than or equal to 𝐾 and less than or equal to 1.98 equals 0.4283, find 𝐾.
We’re told that 𝑍 is a standard normal random variable. So it has a normal distribution with a mean of zero and a standard deviation of one. We’re told that the probability that 𝑍 is between some unknown value 𝐾 and 1.98 is equal to 0.4283. We can visualize this as the area between these two values under the standard normal curve, although at this stage we don’t know whether the value of 𝐾 is positive or negative. So we don’t know which side of the mean it falls on. 𝐾 could be negative, in which case it is in the lower half of the distribution, or 𝐾 could be positive, in which case it is in the upper half of the distribution. Or in fact 𝐾 could be equal to zero. It could be the mean of the distribution.
To work out which of these is the case, let’s first consider the probability that 𝑍 is between zero and this upper limit of 1.98. This will correspond to the area between zero and 1.98 under the standard normal curve. And we can use standard normal distribution tables to find this probability.
Using tables, which give the probability that 𝑍 is greater than or equal to zero and less than or equal to some positive value 𝑧, we find that this probability is equal to 0.4761. But this is greater than 0.4283, which is the value we were given for the probability that 𝑍 is between 𝐾 and 1.98. This tells us that instead of adding on some area for the probability that 𝑍 is between 𝐾 and zero, we actually need to subtract some area. This tells us that 𝐾 must be positive.
Referring to our second diagram then, we have that the area between 𝐾 and 1.98 is 0.4283. And the area between zero and 1.98 is 0.4761. The difference between these two values, which is 0.0478, gives the probability that 𝑍 is between zero and 𝐾. We can then use our standard normal distribution tables in reverse to look up the 𝑧-score, which is the value of 𝐾, associated with a probability of 0.0478. This value is towards the top of the table, and we see that it is associated with a 𝑧-score of 0.12.
So by first determining that the value of 𝐾 was positive and then determining how much of this area was between zero and 𝐾, we were able to use our standard normal distribution tables to show that the value of 𝐾 is 0.12.
