Question

Some IQ tests are standardized to a Normal model ​N(100​,15​). ​a) What cutoff value bounds the highest 10​% of all​ IQs? ​b) What cutoff value bounds the lowest 35​% of the​ IQs? ​c) What cutoff values bound the middle 60​% of the​ IQs?

Answer 1

Answer:

a. 104.969

b. 98,509

c. (96,739;103.261)

Step-by-step explanation:

Hello!

You know that the scores of an IQ test have a normal distribution with μ= 100 and σ²= 15.

To resolve all these questions you need to work under the standard normal distribution and the tables of the distribution.

a.

You need to find the cutoff value that bound the highest 10% of all IQs. This means that you need to find the value from which 10% of the IQ values ​​are found, symbolically:

P(X≥a)= 0.10

If above this value is 10% of the distribution, then below it you can find 90% of the distribution, so you can rewrite it as:

P(X<a)= 0.90

Now you have to standardize it:

P(Z<$\frac{a-Mu}{Sigma }$)= 0.90

Now you look in the distribution table what value of Z corresponds to 0.90 of distribution, note since it accumulates 90% of the distribution, this Z value will be positive.

$Z_{0.90}= 1.283$

$\frac{a-100}{\sqrt{15} } = 1.283$

$a= (1.283*\sqrt{15}) + 100$

a= 104.969

b.

In this item you need to find the cutoff value that bounds the lowest 35%, this means that you are looking for a value that has below it 35% of the distribution and 65% of the distribution is above it, symbolically:

P(X≤b)=0.35

Under the standard normal:

P(Z≤$\frac{b-Mu}{Sigma}$)= 0.35

Now in this case the Z value will be on the left side of the distribution and it will be nehative:

$Z_{0.35}= -0.385$

$\frac{b-100}{\sqrt{15} } = -0.385$

$b= (-0.385*\sqrt{15}) + 100$

b= 98,509

c.

Now you have to find the cutoff values that bound the middle 60% of the IQs. This one is a little tricky let's say that you have an interval with two bonds (c;c') and from the 100% of the distribution you know that 60% is between these two values, then 40% of the distribution is outside them.

The text specifies that it is the middle 60%, in other words, the interval is in the middle of the distribution leaving below c and above c' two tails, if both tails are equal, then the remaining 40% of the distribution is in both tails, 20% each tail.

Now with this in mind, we can say that there is 20% of the distribution below "c" and 20% + 60% below "c'"

It's a little difficult to visualize without a graph so I've attached one with all percentages on it so you can see it while reading the explanation.

Symbolically:

P(c≤X≤c')=0.60

⇒P(X≤c')= 0.80

P(Z≤$\frac{c'-Mu}{Sigma}$)= 0.80

$Z_{0.80}= 0.842$

$\frac{c'-100}{\sqrt{15} }$

$c'= (0.842*\sqrt{15}) + 100$

c'= 103.261

⇒ P(X≤c)= 0.20

The standard normal distribution is symmetrical and centered in 0 since this interval is in the middle of the distribution and the tails are equal, then you can say that the value of  $Z_{c}= - Z_{c'}$ this means that if you look in the table for the value that has 20% of the distribution above you'll find that it is -0.842

$Z_{0.20}= -0842$

$\frac{c-100}{\sqrt{15} } = -0.842$

$c= (-0.842*\sqrt{15}) + 100$

c= 96,739