The education department of state is interested in funding an education assistance program for 12 year-old children whose intelligence quotients lie in the bottom 5% of the population. According to national data, the IQ's of the 12 year-old population are normally distributed with a mean of 96 and a standard deviation of 22.
Calculate the IQ below which a child would be eligible to apply for this education assistance program (X). You may find this standard normal table useful. Give your answer as a whole number.
I don't know what tables you are using, I don't see any.
Instead of tables you can use this
http://davidmlane.com/hyperstat/z_table.html
Use the second graph, fill in 96 for mean, 22 for the standard deviation, .05 for shaded area and click on the below button.
you should get 59.8132
To calculate the IQ below which a child would be eligible to apply for the education assistance program, we need to find the cutoff IQ score corresponding to the bottom 5% of the population.
Since the IQ scores are normally distributed with a mean of 96 and a standard deviation of 22, we can use the standard normal distribution table.
First, let's convert the IQ scores to standard scores (z-scores) using the formula:
z = (X - μ) / σ
where X is the IQ score, μ is the mean (96), and σ is the standard deviation (22).
To find the cutoff z-score for the bottom 5% of the population, we look for the corresponding value in the standard normal table.
From the table, we find that the z-score for the bottom 5% is approximately -1.645.
Now, we can solve for X using the z-score formula:
-1.645 = (X - 96) / 22
Multiply both sides by 22:
-1.645 * 22 = X - 96
Approximately,
-36.19 = X - 96
Add 96 to both sides:
X = 96 - 36.19
X ≈ 59.81
To give the answer as a whole number, we round down to 59.
Therefore, the IQ below which a child would be eligible to apply for this education assistance program is 59.
To find the IQ below which a child would be eligible to apply for the education assistance program, we need to determine the IQ score that corresponds to the bottom 5% of the population.
Given that the IQ scores of 12-year-old children are normally distributed with a mean of 96 and a standard deviation of 22, we can use the z-score formula:
z = (X - μ) / σ
Where:
X is the IQ score we are trying to find
μ is the mean of the IQ scores (96)
σ is the standard deviation of the IQ scores (22)
In this case, since we want to find the IQ score below which 5% of the population falls, we need to find the z-score that corresponds to the 5th percentile.
First, we need to find the z-score using the standard normal distribution table or a statistical calculator. The z-score representing the 5th percentile is approximately -1.645.
Now we can solve for X by rearranging the z-score formula:
-1.645 = (X - 96) / 22
Multiply both sides by 22:
-36.19 = X - 96
Add 96 to both sides:
X = 96 - 36.19
X ≈ 59.81
Therefore, the IQ score below which a child would be eligible to apply for this education assistance program (X) is approximately 60.