Can someone please help me with this? Thank you! Scores on the Stanford-Binet IQ
ID: 3334805 • Letter: C
Question
Can someone please help me with this? Thank you! Scores on the Stanford-Binet IQ Test are normally distributed, with a mean of 100 and a standard deviation of 16. Use this to answer the following questions. a. What percent of the population have IQs above 130? b. What percent of the population have IQs below 70? c. What percent of the population have IQs between 120 and 150? d. What IQ would you need to be in the 90th percentile? e. What IQ would you need to be in the 99th percentile? f. If outliers are defined as values more than 3 standard deviations from the mean, what percent of the population are outliers? What are their IQs? Explain your response in relation to the Empirical Rule Suppose your class of 28 students has an average IQ of 110. Might your class be representative of the general population? In other words, what is the probability that the mean of a random sample of the population would have an IQ at least this large? Support your answer. (H: You need to know the Central Limit Theorem to address this question.) What does this tell you about the students in your class? g.Explanation / Answer
For mean 100 and S.D. of 16 with a normal distribution.
a)
above 130
z score at 130
Z = (X - ) /
Z = (130 - 100) / 16
Z = 1.875
From z table
p for above 130 = (1-(value from z score table))
percantage of population having IQ above 130 = 3.04%
b)below 70
Z = (X - ) /
Z = (70 - 100) / 16
Z = -1.875
From z-score table
percentage at for below 70 IQ is 3.04%
c) between 120 and 150
from z-score table
percantage betwwen 120 to 150 IQ is 10.48%
d) 90 percantile
z score at .9 probability is Z = 1.28155
Z = (X - ) /
1.28155*16 = x-100
x= 100 +20.5048
x= 120.50
90percentile is at 120.50 IQ score