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1 vote
Remember that z=

σ
x−μ

3. IQ tests are designed to be normally distributed with a mean of 100 and a standard deviation of 15 . a. To join Mensa, a special organization for exceptionally bright individuals, you must have an IQ of at least 138 . What percent of the population can join Mensa? b. Many Sociologists consider a child as "gifted" if their IQ is above 150 . i. What percent of the population is considered to be gifted? ii. If a particular school district contains 12,000 students, then how many of them would you expect to be labeled as being gifted? c. What is the probability that a group of 50 individuals has a mean IQ score less than 95 ? Would you consider this an odd event? Why or why not? d. To the nearest whole number, what is the minimum IQ for the top 1% of individuals?

2 Answers

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Final answer:

a. Approximately 0.66% of the population can join Mensa. b. Approximately 0.04% of the population is considered to be gifted and in a school district of 12,000 students, around 5 would be labeled as gifted. c. The probability that a group of 50 individuals has a mean IQ score less than 95 is approximately 0.30%. d. The minimum IQ for the top 1% of individuals is approximately 135.

Step-by-step explanation:

a. To join Mensa:

We know that the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. To find the percentage of the population that can join Mensa (with an IQ of at least 138), we need to find the area under the normal curve to the right of z= (138-100)/15 = 2.53. Using a standard normal table or a calculator, we find that the area to the right of z=2.53 is approximately 0.0066, which is equivalent to 0.66%. Therefore, about 0.66% of the population can join Mensa.

b. Many Sociologists consider a child as "gifted" if their IQ is above 150:

i. To find the percentage of the population considered to be gifted (IQ above 150), we need to find the area under the normal curve to the right of z= (150-100)/15 = 3.33. Using a standard normal table or a calculator, we find that the area to the right of z=3.33 is approximately 0.0004, which is equivalent to 0.04%. Therefore, about 0.04% of the population is considered to be gifted.

ii. To find the number of students in a particular school district expected to be labeled as gifted, we can use the percentage from part b (0.04%) and multiply it by the total number of students in the district (12,000). 0.04% of 12,000 is 4.8, so we would expect around 5 students to be labeled as gifted.

c. Probability of a group of 50 individuals having a mean IQ score less than 95:

To find the probability, we need to calculate the z-score for the mean of 95 in a sample of size 50. The z-score is given by (95-100) / (15/sqrt(50)). Simplifying this expression, we find that the z-score is -2.74.

Using a standard normal table or a calculator, we find that the area to the left of z=-2.74 is approximately 0.0030, which is equivalent to 0.30%. Therefore, the probability that a group of 50 individuals has a mean IQ score less than 95 is 0.30%. This can be considered an odd event since it is a relatively small probability.

d. Minimum IQ for the top 1% of individuals:

To find the minimum IQ for the top 1%, we need to find the z-score that corresponds to an area of 0.01 to the right of it. Using a standard normal table or a calculator, we find that the z-score is approximately 2.33. We can then solve for the minimum IQ by using the formula: z = (x - μ) / σ, where z is the z-score, x is the IQ score, μ is the mean, and σ is the standard deviation. Rearranging the formula, we get x = (z * σ) + μ. Substituting the values, we get x = (2.33 * 15) + 100 = 134.95. Therefore, the minimum IQ for the top 1% of individuals is approximately 135.

User Tim Hope
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3 votes

Final answer:

a. The percentage of the population that can join Mensa is approximately 49.46%. b. i. The percentage of the population considered to be gifted is approximately 0.0429%. ii. In a particular school district with 12,000 students, approximately 5.15 students would be labeled as gifted. c. The probability that a group of 50 individuals has a mean IQ score less than 95 is approximately 11.90%. d. The minimum IQ for the top 1% of individuals is approximately 136.

Step-by-step explanation:

a. To find the percentage of the population that can join Mensa, we need to calculate the z-score for an IQ of 138 based on the normal distribution with a mean of 100 and a standard deviation of 15. Using the formula z = (x - μ) / σ, we have z = (138 - 100) / 15 = 2.53. Looking up this z-score in a standard normal distribution table, we find that the percentage of the population with an IQ of at least 138 is approximately 0.4946, or 49.46%.

b. i. To find the percentage of the population considered to be gifted, we need to calculate the z-score for an IQ of 150. Using the same formula as before, we have z = (150 - 100) / 15 = 3.33. Looking up this z-score, we find that the percentage of the population with an IQ above 150 is approximately 0.000429, or 0.0429%.

ii. To find the number of students in a particular school district who would be labeled as gifted, we need to multiply the percentage calculated in part b.i by the total number of students (12,000). So, the number of students expected to be labeled as gifted is approximately 0.0429% * 12,000 = 5.15 students.

c. To find the probability that a group of 50 individuals has a mean IQ score less than 95, we need to calculate the z-score for an IQ of 95 using the formula z = (x - μ) / (σ / √n), where n is the sample size. Plugging in the values, we have z = (95 - 100) / (15 / √50) = -1.18. Looking up this z-score, we find that the probability is approximately 0.1190, or 11.90%. Whether this is considered an odd event or not depends on the specific context and criteria set by the researcher or organization.

d. To find the minimum IQ for the top 1% of individuals, we need to calculate the z-score corresponding to the top 1% in the standard normal distribution. Looking up this z-score, we find it to be approximately 2.33. Using the z-score formula, we can solve for x: 2.33 = (x - 100) / 15. Rearranging the equation, we have x = 2.33 * 15 + 100 = 135.95. Rounding to the nearest whole number, the minimum IQ for the top 1% of individuals is 136.

User Tranquil Tarn
by
7.8k points
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