In an intelligence test administered to 1000 children, the average score was 42 and fasting the standard deviation 24.Assuming normal distribution, find the number of children with scores between 30 and 60
30 is .5 s.d. below the mean , and 60 is .75 s.d. above the mean
use a z-score table to find the % of the population in the range
To find the number of children with scores between 30 and 60, we need to use the properties of a normal distribution.
First, we can calculate the z-scores for the lower and upper score limits using the formula:
z = (x - μ) / σ
Where:
- x is the score we're interested in (30 and 60, in this case)
- μ is the mean score (42)
- σ is the standard deviation (24)
For the lower score limit (30):
z_lower = (30 - 42) / 24 = -0.5
For the upper score limit (60):
z_upper = (60 - 42) / 24 = 0.75
We can then use a standard normal distribution table (also known as a z-table) to find the proportion of scores between these two z-scores.
The z-table gives us the cumulative probability up to a certain z-score. So, to find the probability of scores below 30, we look up the z-score of -0.5, which gives us a cumulative probability of 0.3085.
Similarly, to find the probability of scores below 60, we look up the z-score of 0.75, which gives us a cumulative probability of 0.7734.
To find the proportion of scores between 30 and 60, we subtract the cumulative probability of scores below 30 from the cumulative probability of scores below 60:
P(30 ≤ x ≤ 60) = P(x ≤ 60) - P(x ≤ 30)
= 0.7734 - 0.3085
= 0.4649
Since we are given that there are 1000 children in total, we can find the number of children with scores between 30 and 60 by multiplying the proportion by the total population:
Number of children = Proportion * Total population
= 0.4649 * 1000
= 464.9
Therefore, approximately 465 children have scores between 30 and 60.