Question

21. A set of scores measuring aggression is normally distributed with a mean equal to 23 and a standard deviation equal to 2.5. Find the proportion: 1. To the left of x = 19.0 2. To the right of x = 25.5 3. Between the mean and x = 19.0 4. To the left of x = 25.5 5. To the right of x = 19.0

Answer 1

We first standardize the distribution to standard normal

1) proportion to the left of x=19 =Area under the standard normal curve between the vertical lines at the corresponding standardized values

= 19-23/2.5 = -1.6

therefore the probability of proportion under the z score of -1.6 we look into the z score table to look for the area under curve values and we get 0.0548 which when multiplied by 100% we get 5.48% of scores to the left of x=19

2) Z score= 25.5-23/2.5 = 1

therefore the probability of proportion under the z score of 1 we look into the z score table to look for the area under curve values and we get 0.8413 which when multiplied by 100% we get 84.13% of scores to the left of x=25.5

but here we want to find the proportion to the right of the value thus

= 1-0.8413 = 0.1587

3) Here we have to find the proportion between mean and 19

so we know that 50% values are under the mean since a normal curve is symmetrical at mean

and we have already found the proportion under 19

therefore the proportion between mean and 19 =

= 0.5-0.0548 = 0.4452

4) For x=25.5 we have already found the proportion to the left of it as 0.8413 in question 2

for x=19 we have found out the proportion to the left of it as 0.0548

thus for the right of it, we simply subtract it from 1

that is = 1-0.0548 = 0.9452, which is 94.52%