Let z be a normal random variable with mean 0 and standard deviation 1. The 4th decile of z is ___.
a) 0.67 b) –1.25 c) -0.25 d) 1.28 e) 0.50
To find the 4th decile of a normal random variable, we need to determine the value below which 40% of the data falls.
First, let's calculate the z-score corresponding to the 4th decile, which is the value below which 40% of the data falls. To do this, we use the standard normal distribution table (also known as the z-table) or a statistical calculator.
The z-score corresponding to the 4th decile is -0.253. Let's find out how to calculate it:
1. Find the z-score associated with the cumulative probability of 0.40. Using the standard normal distribution table, we find that the closest value is 0.25. Since the distribution is symmetrical about zero, we take the negative z-score.
2. However, we need to account for the fact that the 4th decile is not exactly at 0.40, but slightly below it. To do this, we use linear interpolation. Since our z-score of 0.25 is associated with a cumulative probability of 0.5987, we can calculate the additional contribution by subtracting 0.40 from 0.5987 to get 0.1987.
3. We divide this additional contribution by the difference between the cumulative probabilities corresponding to the z-scores on either side of 0.40. These are 0.5987 and 0.5019, resulting in a difference of 0.0968.
4. Finally, we multiply the difference by the difference in z-scores on either side of 0.40 (0.25 and 0.24) to get the additional contribution in terms of z-score. This gives us (0.0968 * 0.01) = 0.000968.
5. Subtracting this additional contribution from our initial z-score, we get -0.25 - 0.000968 = -0.250968. Rounding to three decimal places, this gives us a final z-score of -0.251.
The 4th decile of the normal random variable is approximately -0.251.
Therefore, the correct answer is c) -0.25.