Question

In: Statistics and Probability

Most tests that give IQ scores are designed to have a mean of 100 and a...

Most tests that give IQ scores are designed to have a mean of 100 and a standard deviation of 15. IQ testing is one way in which people are categorized as having different levels of intellectual disability (the category formerly known as mental retardation). Prior to DSM-V, when the term mental retardation was used, four categories were distinguished. Two of those categories are shown below:

-IQ of 20-35 (in combination with impaired life skills): severe mental retardation

-IQ of 50-70 (in combination with impaired life skills): mild mental retardation

People with scores above 70 were not considered to have mental retardation.

The IQ test was used first, and then, if a person’s score fell into the range for mental retardation (70 or below) above, that person would be assessed for life skill impairments.

What percentage of people would be expected to have IQ scores that might put them into the severe mental retardation range? Assume that IQ scores are normally distributed in the population. (Enter a number without the percent sign, rounded to the nearest 2 decimal places)

Solutions

Expert Solution

Here we want to find the percentage of people would be expected to have IQ scores that might put them into the severe mental retardation range.

From the given information the range of severe mental retardation range is 20 to 35.

Let's say that X ~ N ( = 100 , = 15)

So, here we want to find P( 15 < X < 35 ) = P( X < 35 ) - P(X < 15) ........( 1 )

Let's use excel:

P(X < 35 ) = "-NORMDIST(35,100,15,1)" = 0.00000734342

P(X < 15) = "-NORMDIST(20,100,15,1)" =  0.000000048213

Plug these values in equation ( 1 ) , we get,

P( 20 < X < 35 ) = 0.00000734342 -  0.000000048213

= 0.00000729521= 0.000729521% = 0.00% (after rounding upto two decimal places.

If the person's score is less than 70 then the chance of that person in the range of severe mental retardation range is as follows:

p = P(20 < X < 35)/ P(X < 70) ...( 2 ).

Let P(X < 70) = "=NORMDIST(70,100,15,1)" = 0.02275

therefore p = 0.00000729521 / 0.02275 = 0.032067 = 0.03% (rounding upto two decimal places).

So the person whose score is less than or equal to 70 and he is in the range of severe mental retardation range is 0.03%


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