The middle 69% of a normally distributed population lies between what two standard scores? (Give your answers correct to two decimal places.)0.31 and 1.69 was my answers and they are wrong????. .
You want the Z values containing 69%. Those are -1 <=Z <= 1
(well, 1.015, really)
that is, within 1 std of the mean.
To find the standard scores that correspond to the middle 69% of a normally distributed population, you'll need to use the standard normal distribution table (also known as the Z-table). Here's how you can calculate it:
1. Start by finding the proportion of the middle 69% in decimal form: 69% divided by 100%. This gives us 0.69.
2. Next, find the proportion in the tails (outside of the middle 69%). Since the normal distribution is symmetrical, we divide this proportion by 2: 1 - 0.69 = 0.31 / 2 = 0.155.
3. Now you need to find the corresponding Z-scores for the tail proportions. Use the Z-table to find the Z-score that corresponds to the tail proportion of 0.155. Looking up this value in the Z-table, you'll find that it corresponds to approximately -0.99.
4. Finally, calculate the Z-scores for the middle 69%. Subtract the Z-score for the tail proportion from 1 to find the Z-score for the upper bound, and add the Z-score for the tail proportion to -1 to find the Z-score for the lower bound.
Upper bound Z-score = 1 - (-0.99) = 1.99
Lower bound Z-score = -1 + (-0.99) = -1.99
Correct to two decimal places, the middle 69% of a normally distributed population lies between approximately -1.99 and 1.99 standard scores.