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.