Question

Suppose a distribution of 400 items has mean 125 and standard deviation of 5. Find the following.

a) z-score of x=133 b) z-score of x=113 c)How many items are less than first quartile?

Answer 1

The z-score for x=133 is 1.6 and for x=113 it is -2.4. To determine the number of items less than the first quartile in a normal distribution, we would expect 25% of the items, which amounts to 100 out of 400 items.

Step-by-step explanation:

To calculate the z-score of a given value x, we use the formula:

z = (x - mean) / standard deviation

a) For x = 133:

z = (133 - 125) / 5 = 8 / 5 = 1.6

b) For x = 113:

z = (113 - 125) / 5 = -12 / 5 = -2.4

To determine how many items are less than the first quartile, we need to know the z-score corresponding to the first quartile in a normal distribution, which is approximately -0.675 (25th percentile). Then, we apply the following:

Q1 = mean + (z * standard deviation)

We calculate the first quartile value and use the cumulative properties of the normal distribution to estimate the number of items. Since 25% of the data lies below the first quartile, and we have 400 items, we expect that 0.25 * 400 = 100 items are less than the first quartile.