Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Remember that a sample distribution describes a statistic of a population, like

ID: 3172182 • Letter: R

Question

Remember that a sample distribution describes a statistic of a population, like the mean. From a random sample of size 50, the bar x = 15/x If the population has mu = 17 and sigma = 2. then the sampling distribution of bar x is a mean of 17 and standard deviation of 0.282843 (use 4 decimals in your answer). The time (in days) that child needs to fight the flu virus depends on the child's immunity and the amount of medicine available. The mean number of days to get over the flue for a child is mu = 2.3 days, but some children take longer. The distribution of fighting the flue is skewed right due to children that lack medicine and have poor immune systems, with a standard deviation of sigma = 1.0. What is the probability that the average child has the flu for longer then 2 days? If we have a sample of 100 children that have the flu, what is the mean and standard deviation for the sampling distribution? Mean = 2.3 standard deviation = 0.1 (use 4 decimals in your answers) Using those numbers in your normal curve, and your TI, determine the probability of an average child having the flu for longer then 2 days. 0.13%

Explanation / Answer

Solution:

We know that the estimate for the mean for sampling distribution is given as the population mean and the estimate for standard deviation for the sampling distribution is given as the standard error.

Estimate for standard deviation = /sqrt(n)

Where n is sample size

Estimate for standard deviation = 2/sqrt(50)

Estimate for standard deviation = 0.2828

Question 4

We are given n = 100

µ = 2.3

= 1.0

Mean for sampling distribution = 2.3

Standard deviation for sampling distribution = /sqrt(n) = 1.0/sqrt(100) = 1/10 = 0.1000

Standard deviation for sampling distribution = 0.1000

Now, we have to find P(X>2)

P(X>2) = 1 – P(X<2)

Z score for X = 2

Z = (X – mean) / SD

Z = (2 – 2.3) / 0.1 = 0.3/0.1 = 3

P(X<2) = P(Z<3) = 0.99865

P(X>2) = 1 – P(X<2) = 1 – 0.99865 = 0.00135

Answer: 0.00135 or 0.14%