Scores on the Stanford Binet Intelligence Scale (SBIS) are distributed as normal with a mean of 100 and a standard deviation of 16. Use the probabilities under the Normal Curve to answer the following questions.
a. A child scored 84 on the SBIS. What is his z score? -1
b. What percentage of children will fall between 68 and 116?
c. What percentage of children will fall below 100?
d. A child scored 132 on the SBIS. What is her z score? 2
To answer these questions, we need to use the standard score or z-score formula:
z = (x - mean) / standard deviation
where x is the given score, mean is the mean of the distribution, and the standard deviation is the standard deviation of the distribution.
a. To find the z-score for a score of 84:
z = (84 - 100) / 16 = -1
Therefore, the z-score for a score of 84 is -1.
b. To find the percentage of children who will fall between 68 and 116, we need to find the area under the normal curve between these two z-scores.
First, we find the z-scores for 68 and 116:
z_68 = (68 - 100) / 16 = -2
z_116 = (116 - 100) / 16 = 1
We can then use a standard normal distribution table or a calculator to find the percentage of children between these two z-scores. From the table or calculator, we find that the area to the left of -2 on the standard normal distribution is 0.0228 and the area to the left of 1 is 0.8413.
The percentage of children falling between z = -2 and z = 1 is:
0.8413 - 0.0228 = 0.8185
Therefore, approximately 81.85% of children will fall between scores of 68 and 116.
c. To find the percentage of children falling below 100, we need to find the area to the left of the z-score of 0, which represents the mean of the distribution.
Using the z-score formula, we find:
z = (100 - 100) / 16 = 0
The area to the left of z = 0 on the standard normal distribution table is 0.5000.
Therefore, approximately 50% of children will fall below a score of 100.
d. To find the z-score for a score of 132:
z = (132 - 100) / 16 = 2
Therefore, the z-score for a score of 132 is 2.