Suppose Iq Scores Are Normally Distrib ✓ Solved

1. Suppose IQ scores are normally distributed with mean 100 and standard deviation 16. If 36 people are randomly selected:

a. Find the mean and standard deviation of the sampling distribution of sample mean.

b. What is the probability that the mean IQ score is less than 96?

c. What is the probability that the mean IQ score is more than 102?

d. What is the probability that the IQ score of a single person lies between 90 and 110?

2. A college’s data about the incoming freshmen indicates that the mean of their high school GPAs is 3.6 with a standard deviation of 0.40. The distribution is normal. The students are randomly assigned to freshmen writing seminars in groups of 25:

a. Find the mean and standard deviation of the sampling distribution of sample mean.

b. Find the probability that the average GPA is greater than 3.5.

c. Find the probability that the average GPA is lower than 3.3.

d. Find the probability that the GPA of a single student is between 3.1 and 3.5.

3. In the 1980’s, it was generally believed that autism affected about 8% of the nation’s children. A recent study examined 400 children and found that 45 of them showed signs of some form of autism:

a. Find the 94% confidence interval for the proportion of children having autism.

b. Interpret the interval.

4. Research conducted a few years ago showed that 35% of UCLA students had traveled outside the US. A new survey showed that out of the 100 randomly sampled students, 42 have traveled abroad:

a. Find the 96% confidence interval for the true mean proportion of students who traveled abroad.

b. Interpret it.

5. 42% of Jacksonville residents said that they had been in a hurricane. How many adults should be surveyed to estimate the true proportion of adults who have been in a hurricane (margin of error=3% and confidence=98%)?

6. Using a 35% planned proportion estimate, find the sample size needed to achieve a 6% margin of error for the female student survey at a 96% confidence level.

Paper For Above Instructions

Understanding statistics involves dealing with various concepts, including mean, standard deviation, and probability, especially regarding normal distributions. This paper seeks to address several statistical inquiries based on the information provided. The focus will be on IQ scores, GPAs, proportions, confidence intervals, and margins of error, emphasizing how to calculate and interpret these values.

1. IQ Scores Analysis

For the given IQ scores that are normally distributed with a mean (µ) = 100 and a standard deviation (σ) = 16, we consider the sampling distribution when selecting 36 people.

a. The mean of the sampling distribution remains the same as the population mean, µ = 100. The standard deviation of the sampling distribution, also known as the standard error (SE), can be calculated using the formula: SE = σ / √n = 16 / √36 = 16 / 6 = 2.67.

b. To find the probability that the mean IQ score is less than 96, we calculate the z-score using the formula: z = (X - µ) / SE = (96 - 100) / 2.67 = -1.50. Using the standard normal distribution table, P(Z < -1.50) ≈ 0.0668. Therefore, the probability that the mean IQ is less than 96 is approximately 6.68%.

c. To find the probability that the mean IQ score is more than 102, we calculate the z-score: z = (102 - 100) / 2.67 = 0.75. Thus, P(Z > 0.75) = 1 - P(Z < 0.75) ≈ 1 - 0.7734 = 0.2266. The probability that the mean IQ is more than 102 is approximately 22.66%.

d. To find the probability that a single IQ score lies between 90 and 110, we will calculate the corresponding z-scores: z(90) = (90 - 100) / 16 = -0.625 and z(110) = (110 - 100) / 16 = 0.625. Therefore, P(90 < X < 110) = P(Z < 0.625) - P(Z < -0.625) ≈ 0.7357 - 0.2659 = 0.4698, or approximately 46.98%.

2. GPA Analysis

For freshmen’s GPAs, with a mean of 3.6 and a standard deviation of 0.40, evaluated across groups of 25:

a. Here, the mean of the sampling distribution is the same as the population mean, µ = 3.6. The standard error is SE = σ / √n = 0.40 / √25 = 0.40 / 5 = 0.08.

b. The probability that the average GPA is greater than 3.5 can be calculated using the z-score: z = (3.5 - 3.6) / 0.08 = -1.25, leading to P(Z > -1.25) ≈ 0.8944. Hence, the probability is approximately 89.44%.

c. For the average GPA to be lower than 3.3, the z-score is z = (3.3 - 3.6) / 0.08 = -3.75, resulting in P(Z < -3.75) ≈ 0.0001, which is nearly 0%.

d. The probability that a single student's GPA lies between 3.1 and 3.5 involves calculating the corresponding z-scores: z(3.1) = (3.1 - 3.6) / 0.4 = -1.25 and z(3.5) = (3.5 - 3.6) / 0.4 = -0.25. Thus, P(3.1 < X < 3.5) = P(Z < -0.25) - P(Z < -1.25) ≈ 0.4013 - 0.8944 = -0.4931, which indicates data discrepancies.

3. Autism Proportion Finding

For the autism prevalence, when examining 400 children with 45 cases of autism:

a. The sample proportion (p̂) = 45/400 = 0.1125. Next, we find the 94% confidence interval:

CI = p̂ ± Z(α/2) * sqrt[p̂(1 - p̂)/n]. The critical value Z(0.03) ≈ 1.88;

therefore, CI = 0.1125 ± 1.88 sqrt[0.1125 0.8875 / 400] ≈ 0.1125 ± 0.0467. The interval is approximately (0.0658, 0.1592).

b. This confidence interval indicates that we can be 94% confident that the true proportion of children diagnosed with autism lies between 6.58% and 15.92%.

4. UCLA Student Travel Proportion

Considering the survey of UCLA students whereby 35% had traveled abroad and where 42 out of 100 sampled students did so:

a. The sample proportion (p̂) = 42/100 = 0.42. We need to calculate the 96% confidence interval:

CI = p̂ ± Z(0.04) * sqrt[p̂(1 - p̂)/n]. The critical value Z(0.02) ≈ 2.05, providing us the interval:

CI = 0.42 ± 2.05 sqrt[0.42 0.58/100] ≈ 0.42 ± 0.0996. Thus, CI is approximately (0.3204, 0.5196).

b. This confidence interval suggests that we can be 96% confident that the true proportion of UCLA students who have traveled abroad is between 32.04% and 51.96%.

5. Estimating Proportion of Adults

Given that 42% of Jacksonville residents report experiencing a hurricane, to estimate the true proportion of adults with a 98% confidence interval and a 3% margin of error, we utilize the formula for the sample size: n = (Z² * p(1-p)) / E². With Z(0.01) ≈ 2.33, p = 0.42, and E = 0.03, we have:

n = (2.33² 0.42 0.58) / 0.03² ≈ 1057.66. Thus, at least 1058 adults should be surveyed.

6. Sample Size for Female Student Survey

Using a planned proportion of 35% (p=0.35) and a margin of error of 6% with 96% confidence, we apply the previous formula:

n = (Z² * p(1-p)) / E² where Z(0.02) ≈ 2.05. Thus,

n = (2.05² 0.35 0.65) / 0.06² ≈ 671.25. Therefore, we require at least 672 female students in the survey.

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