I figured out the first question but don't seem to understand the other two
Suppose that the age of students at George Washington Elementary school is uniformly distributed between 6 and 10 years old. 43 randomly selected children from the school are asked their age. Round all answers to 4 decimal places where possible.
What is the distribution of X ? X ~ U =(6 , 10)
Suppose that 43 children from the school are surveyed. Then the sampling distribution is
What is the distribution of ¯x? ¯x ~ N( , )
What is the probability that the average of 43 children will be between 8 and 8.3 years old?
To answer the first question, we need to understand the concept of a uniform distribution. In a uniform distribution, all values within a given range are equally likely to occur. In this case, the age of students at George Washington Elementary school is uniformly distributed between 6 and 10 years old.
The distribution of X, which represents the age of a randomly selected student, is given as X ~ U(6, 10). The notation U(a, b) represents a uniform distribution between a and b.
Now, let's move on to the second question. We are interested in the distribution of the sample mean, ¯x, of 43 children from the school. The sampling distribution of the mean refers to the distribution of all possible sample means that can be obtained from a population.
Given that the age of students is uniformly distributed between 6 and 10 years old, the central limit theorem tells us that as the sample size increases, the sampling distribution of the mean approaches a normal distribution. In this case, with a sample size of 43, we can assume that the distribution of ¯x is approximately normal.
The distribution of ¯x is given as ¯x ~ N(μ, σ/√n), where N represents the normal distribution, μ represents the mean of the population, σ represents the standard deviation of the population, and n represents the sample size.
Since the age of students is uniformly distributed between 6 and 10, we can calculate the mean (μ) and standard deviation (σ) of the population. The mean of a uniform distribution is given by (a + b) / 2, and the standard deviation is given by √[(b - a)^2 / 12].
In this case, a = 6 and b = 10. Therefore, the mean (μ) of the population is (6 + 10) / 2 = 8, and the standard deviation (σ) is √[(10 - 6)^2 / 12] ≈ 1.1547.
Now, let's move on to the final question. We need to calculate the probability that the average (¯x) of 43 children will be between 8 and 8.3 years old.
To find this probability, we need to calculate the z-scores corresponding to the lower and upper limits of the desired range. The z-score formula is given by z = (x - μ) / (σ/√n), where x represents the value, μ represents the mean, σ represents the standard deviation, and n represents the sample size.
For the lower limit, z1 = (8 - 8) / (1.1547/√43) ≈ 0. For the upper limit, z2 = (8.3 - 8) / (1.1547/√43) ≈ 0.985.
Using a standard normal distribution table or a calculator, we can find the corresponding probabilities for these z-scores. Then, we can subtract the probability corresponding to the lower limit from the probability corresponding to the upper limit to find the probability that the average of 43 children will be between 8 and 8.3 years old.