A large computer chip manufacturing plant under construction in Westbank is expe

Question

A large computer chip manufacturing plant under construction in Westbank is expected to add 1400 children to the county's public school system once the permanent work force arrives. Any child with an IQ under 80 or over 135 will require individualized instruction that will cost the city an additional $1750 per year. How much money should Westbank anticipate spending next year to meet the needs of its new special ed students? Assume that IQ scores are normally distributed with a mean of 100 and a standard deviation of 16.

Explanation / Answer

Background information: Essentially what you needto find out is the probability of x amount of students having an IQless than 80 and greater than 135. To find this you need to make astandard normal distribution. You can take a normal distributionsuch as IQ scores and convert it to standard normal form. The areaunder the curve is the probability that a student is in the range80 - 135. You can then take that probability and use the complementrules 1 - p to find the probability of a studentnot being in the range 80 - 135. Step 1: Standardize values given: Use this formula: (x - )/ where x is the value given, is the population mean and is the standard deviation. Z1 = (80 - 100) / (16) = -1.25 Z2 = (135 -100) / (16) = 2.1875 rounded to 2.19 Step 2: Obtain the probability of a student notbeing in the 80 - 135 You need to look these values up in a z-score table: The area under the curve of 2.19 in a normal distribution is:.9957 The area under the curve of -1.25 in a normal distribution is:.1056 To get the area in between the two z scores take Z2.5 - Z-2.5, so,.9957-.1056= .8901 This means that 89.01% of students fall within the 80-135 IQ range.To find the student that don't subtract .8901 from 1 to get .1099.So 10.99% students do not fall within the 80-135IQ range. Step 3: Interpret the answer: Since the school knows that 1400 children will be entering theschool system, about 1400*.1099 = 153.86 rounded down to 153students will need to be accommodated for. So 153*1750 = $267,750will be needed to support these special ed students. Note: I solved a similar problem like this in the forum. For extrapractice and better understanding you can take a look at the posttitled "Normal Distribution". If you need anymore help let me know.

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