In an intelligence test administered to 1000 childrens the average score is 42 and standard deviation 24.find (i)the number children exceeding the score 60.
(ii) the number of the children with score lying between 20 and 40.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to your Z scores, then convert to number.
To find the number of children exceeding the score of 60, we can use the Z-score formula and the given average and standard deviation. The Z-score formula is:
Z = (X - μ) / σ
Where:
Z = Z-score
X = Individual score
μ = Mean (average)
σ = Standard deviation
Step 1: Calculate the Z-score for a score of 60:
Z = (60 - 42) / 24
Z = 0.75
Step 2: Look up the area under the normal distribution curve for a Z-score of 0.75 in a Z-table. The area corresponds to the percentage of scores above 60.
Using a Z-table, the area to the left of Z = 0.75 is 0.7734. This means that 77.34% of scores are below 60 when the average is 42 and the standard deviation is 24.
Step 3: Calculate the number of children exceeding the score of 60:
Number of children = Total number of children * (1 - Area)
Number of children = 1000 * (1 - 0.7734)
Number of children = 226.6
So, there are approximately 227 children who exceed the score of 60.
To find the number of children with scores lying between 20 and 40, we can again use the Z-score formula.
Step 1: Calculate the Z-score for a score of 20:
Z₁ = (20 - 42) / 24
Z₁ = -0.92
Step 2: Calculate the area under the normal distribution curve for a Z-score of -0.92:
Using a Z-table, the area to the left of Z = -0.92 is 0.1808. This means that 18.08% of scores are below 20 when the average is 42 and the standard deviation is 24.
Step 3: Calculate the Z-score for a score of 40:
Z₂ = (40 - 42) / 24
Z₂ = -0.08
Step 4: Calculate the area under the normal distribution curve for a Z-score of -0.08:
Using a Z-table, the area to the left of Z = -0.08 is 0.4681. This means that 46.81% of scores are below 40 when the average is 42 and the standard deviation is 24.
Step 5: Calculate the number of children with scores between 20 and 40:
Number of children = Total number of children * (Area₂ - Area₁)
Number of children = 1000 * (0.4681 - 0.1808)
Number of children = 287.3
So, there are approximately 287 children with scores lying between 20 and 40.
To find the number of children exceeding the score of 60, we first need to convert the scores to z-scores. A z-score is a measure of how many standard deviations a data point is away from the mean.
Using the formula:
z = (x - μ) / σ
where:
z is the z-score,
x is the raw score,
μ is the mean, and
σ is the standard deviation.
To find the z-score for a score of 60:
z = (60 - 42) / 24
z = 0.75
Now, we need to find the area under the normal distribution curve to the right of this z-score. This gives us the proportion of children with scores exceeding 60.
Using a standard normal distribution table or a calculator, we can find that the area to the right of 0.75 is approximately 0.2266.
Finally, to find the number of children exceeding the score of 60, we multiply the proportion by the total number of children:
Number of children exceeding 60 = Proportion * Total number of children
Number of children exceeding 60 = 0.2266 * 1000
Number of children exceeding 60 ≈ 227
Therefore, approximately 227 children exceeded the score of 60.
To find the number of children with scores lying between 20 and 40, we first need to find the z-scores for these values.
For a score of 20:
z = (20 - 42) / 24
z = -0.92
For a score of 40:
z = (40 - 42) / 24
z = -0.0833
Now, we need to find the area under the normal distribution curve between these two z-scores. This gives us the proportion of children with scores lying between 20 and 40.
Using a standard normal distribution table or a calculator, we can find that the area to the left of -0.92 is approximately 0.1788 and the area to the left of -0.0833 is approximately 0.4656.
To find the area between these two z-scores, we subtract the smaller area from the larger area:
Area between -0.0833 and -0.92 = 0.4656 - 0.1788
Area between -0.0833 and -0.92 ≈ 0.2868
Finally, we multiply this proportion by the total number of children:
Number of children with scores between 20 and 40 = Proportion * Total number of children
Number of children with scores between 20 and 40 = 0.2868 * 1000
Number of children with scores between 20 and 40 ≈ 287
Therefore, approximately 287 children have scores lying between 20 and 40.