Respuesta :
Answer:
The IQ score separating the bottom 25% from the top 75%. is 87.3
Step-by-step explanation:
Given that adults have IQ scores that are normally distributed with a mean of 98.8 and a standard deviation 17.1.
To find the first quartile, we calculate X such that P(X<z)=0.25.
From the normal distribution table the The z-value that corresponds to an area of 0.25 is z=-0.675
We use the formula:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
We substitute to get:
[tex]-0.675=\frac{X-98.8}{17.1}[/tex]
This implies that:
[tex]17.1*-0.675=X-098.8[/tex]
[tex]-11.5425=X-098.8[/tex]
Solve for x to get:
[tex]x=98.8-11.5425[/tex]
X=87.2575
Using the normal distribution, it is found that the IQ score separating the bottom 25% from the top 75% is of 87.
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Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula, which in a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- Each z-score has a p-value associated, which we find looking at the z-table, and this p-value represents the percentile of the measure X.
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- Mean of 98.8 means that [tex]\mu = 98.8[/tex]
- Standard deviation of 17.1 means that [tex]\sigma = 17.1[/tex];
- The value of the interest is the 25th percentile, which is X when Z has a p-value of 0.25, so X when Z = -0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 98.8}{17.1}[/tex]
[tex]X - 98.8 = -0.675(17.1)[/tex]
[tex]X = 87[/tex]
The IQ score is 87.
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