The weschler intelligence scale iq for children is...

The Weschler Intelligence Scale (IQ) for Children is approximately normally distributed with a mean of 100 and a standard deviation of 15. a) What is the probability that a randomly selected child has an IQ score above 120? b) What is the probability that a randomly selected child has an IQ score between 93 and 110? c) What IQ score would place a child in the top 3% of all IQ scores? d) What would be the range of scores that represent the middle 90% of all children's IQ score?

Transcript

00:01 For this problem, the iq for children is normally distributed with a mean of 100 and a standard deviation of 15.

00:12 So we define a random variable x and that is equal to the iq scores for children.

00:20 So for the first question, we are to determine the probability that a randomly selected child has an iq score above 120.

00:29 So we can write that as the probability that x is greater than 100.

00:34 Now since this distribution follows a normal distribution we need to compute the z score so the formula for computing the z score is giving us z is equal to x minus the mean divided by the standard deviation so this becomes probability that x becomes z that is greater on 120 minus the mean just 100 divided by the standard deviation and that becomes probability that z is greater than 1 .33.

01:30 In computing this probability, this becomes 1 minus the probability that z is less than 1 .033 and this is obtained from the standard z table and we have 0 .9082.

02:00 So we do the subtraction and we have 0 .0918.

02:14 So for the next question we are to determine the probability that a randomly selected child has an iq score between 93 and 110.

02:25 So you can write that as the probability that 93 is less than or equal to x.

02:32 That is less than or equal to 110.

02:36 So we compute the z scores as well.

02:40 You have 93 minus 100 divided by 15 that is less than or equal to z which is also less than or equal to 110 minus 100 then he divided by 15 so becomes negative 0 .47 less than or equal to z and that is less than or equal to 0 .67 so this becomes the normal of 0 .67 minus the normal of negative 0 .47.

03:47 Now this gives us 0 .786 minus 0 .3192.

03:58 And these values are obtained from the z table.

04:02 We do the subtraction and we have 0 .429, sorry, 0 .4294.

04:14 Move on to the next question.

04:23 We are to determine the iq score that would place a child in the top 3 % of all iq scores.

04:30 So we want to present this on a standard normal curve.

04:39 We are going to assume that this region covers the top 3%.

04:48 There are 3%, which is 4 to 0 .03...