In order to find the area under the standard normal curve to the left of a specified z-score, we will need to use the cumulative distribution function (CDF) of the standard normal distribution. The CDF of a random variable is a function which measures the probability that a random variable will take on a value less than or equal to some given value.
Let's find the area to the left of z = 2.19 using the cumulative distribution function.
The z-score of 2.19 represents the number of standard deviations that an element is away from the mean of the distribution. Given that the z-distribution is symmetrical and the total area under the curve is equal to 1, the area to the left of the mean (z=0) is exactly 0.5. As the value of z increases, so does the area to the left of it.
For a z-score of 2.19, we look up this value in the z-distribution table or use a statistical software or calculator to find the cumulative area. For z=2.19, the value will be approximately 0.9857. Thus, about 98.57% of data falls below a z-score of 2.19.
Now to find the z-scores using the inverse of the cumulative distribution function, we use the specified probabilities. Note: the inverse of the cumulative distribution function gives us the z-score.
a) For a left-tail probability of 0.9750, using the z-distribution table or a statistical software or calculator, we find that the z-score that corresponds to this probability is approximately 1.96.
b) For a left-tail probability of 1.645, we are actually looking for a z-score. There is a confusion in the question because probabilities range between 0 and 1. However, the z-score that corresponds to a probability of 0.95 (where usually 1.645 is found), is indeed 1.645.
c) For a left-tail probability of 0.0250, using the z-distribution table, we find that the z-score that corresponds to this probability is approximately -1.96.
d) For a left-tail probability of -1.645, we once again see there is a confusion in the question. Again, a z-score is asked for, and if we are to consider this as an inverse operation, with this z-score the corresponding left-tail probability is approximately 0.05.