. An intelligence test used in a particular country has scores which are normally distributed with
mean 100 and standard deviation 15. In a randomly selected group of 500 people sitting the test,
estimate how many have a score
(i) higher than 140
(ii) below 120
(iii) between 100 and 110
(iv) between 85 and 90.
To estimate the number of people in each group, we will use the properties of the normal distribution. The normal distribution is characterized by its mean and standard deviation. In this case, we know that the mean score is 100 and the standard deviation is 15.
(i) To estimate how many people have a score higher than 140, we need to find the proportion of the population with a score above 140. We can then multiply this proportion by the total number of people in the group (500) to get an estimate of the number of individuals.
To find the proportion, we will use a Z-score calculation. The Z-score tells us how many standard deviations a given value is away from the mean. In this case, we want to find the proportion of scores above 140, so we calculate the Z-score for 140 using the formula:
Z = (X - μ) / σ
where X is the score (140), μ is the mean (100), and σ is the standard deviation (15).
Z = (140 - 100) / 15
Z = 40 / 15
Z ≈ 2.67
Using a Z-table or a calculator, we can find the proportion corresponding to a Z-score of 2.67. Looking up the value in the table, we find that the proportion is approximately 0.9968.
Therefore, we estimate that approximately 0.9968 * 500 = 498.4 people would have a score higher than 140. Since it is not possible to have a fraction of a person, we can round this estimate to the nearest whole number. So, we can estimate that about 498 people have a score higher than 140.
(ii) To estimate how many people have a score below 120, we will use the same method. We calculate the Z-score for 120:
Z = (120 - 100) / 15
Z = 20 / 15
Z ≈ 1.33
Using the Z-table or a calculator, we find that the proportion corresponding to a Z-score of 1.33 is approximately 0.908.
Therefore, we estimate that approximately 0.908 * 500 = 454 people would have a score below 120.
(iii) To estimate how many people have a score between 100 and 110, we need to find the proportion between these two scores.
First, we calculate the Z-score for 100 and 110:
Z1 = (100 - 100) / 15 = 0
Z2 = (110 - 100) / 15 = 2/3 ≈ 0.67
Using the Z-table or a calculator, we find that the proportion corresponding to a Z-score of 0 is 0.5, and the proportion corresponding to a Z-score of 0.67 is approximately 0.748.
To find the proportion between these two scores, we subtract the proportion corresponding to the lower Z-score (0.5) from the proportion corresponding to the higher Z-score (0.748):
Proportion = 0.748 - 0.5 = 0.248
Therefore, we estimate that approximately 0.248 * 500 = 124 people would have a score between 100 and 110.
(iv) To estimate how many people have a score between 85 and 90, we use the same method. We calculate the Z-scores for 85 and 90:
Z1 = (85 - 100) / 15 = -1
Z2 = (90 - 100) / 15 = -2/3 ≈ -0.67
Using the Z-table or a calculator, we find that the proportion corresponding to a Z-score of -1 is approximately 0.1587, and the proportion corresponding to a Z-score of -0.67 is approximately 0.2514.
To find the proportion between these two scores, we subtract the proportion corresponding to the lower Z-score (0.1587) from the proportion corresponding to the higher Z-score (0.2514):
Proportion = 0.2514 - 0.1587 = 0.0927
Therefore, we estimate that approximately 0.0927 * 500 = 46.35 people would have a score between 85 and 90. Rounded to the nearest whole number, we can estimate that about 46 people have a score between 85 and 90.