What is the variance of the distribution of the average an IID draw of n observations from a population with mean μ and variance σ2.
The avarage of random samples from a population is itself a random variable having a distribution. This distribution is centered around the poplulation mean. The variance of the sample mean is σ2 / n.
Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. About what is the probability that a random 35-44 year old has a DBP less than 70?
Given that DBPs is normally distributed, N(80,10^2) . That probability that a random 35-40 year old has a DBP less than 70 is
pnorm(70,mean=80, sd=10)## [1] 0.1586553Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?
Brain volume for adult women is N(1100,75^2) . We use qnorm to calculate the 95th percentile.
qnorm(0.95,mean=1100, sd=75)## [1] 1223.364This could also have been calculated by shifting and scaling normals.
We can transform normal random variables to standard normals and vice versa. If X ~ N(μ,σ2) then: Z = (X- μ) / σ ~ N (0,1) . If Z is standard normal, then X= μ + σ*Z ~ N(μ,σ2)
In this case, the answer is μ + σ* (1.645) as 1.645 is 95th quantile of standard normal.
1100 + 75* 1.645 ## [1] 1223.375Refer to the previous question. Brain volume for adult women is about 1,100 cc for women with a standard deviation of 75 cc. Consider the sample mean of 100 random adult women from this population. What is the 95th percentile of the distribution of that sample mean?
Since brain volume is X ~ N(1100,75^2) , The distribution of the sample mean is ~ N (1100, 75^2/100)
qnorm(0.95,mean=1100, sd=75/sqrt (100))## [1] 1112.336You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?
The flip of a fair coin is binomial distribution with Probability=0.5
round (pbinom(3 , prob = 0.5, size =5, lower.tail= FALSE ) * 100,1) ## [1] 18.7The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample mean RDI of 100 people is between 14 and 16 events per hour?
Since we are interested in the distibution of sample mean, We need tha sample error
RDI_mean <- 15
RDI_sd <- 10
RDI_n <- 100
RDI_sample_err <- 10/sqrt(100)Our sample error is 1.
Even if the individual observations of a population with Normal distribution are not normally distributed, The Central limit theorem tells us that the means of the samples will be normally distributed.
The range we’re interested is excatly one standard error away from the mean, hence 68% of the sample means will be within this range.
Consider a standard uniform density. The mean for this density is .5 and the variance is 1 / 12. You sample 1,000 observations from this distribution and take the sample mean, what value would you expect it to be near?
The sample mean would be near the popular mean of standard uniform density.
The number of people showing up at a bus stop is assumed to be Poisson with a mean of 5 people per hour. You watch the bus stop for 3 hours. About what’s the probability of viewing 10 or fewer people?
The poission distribution is useful for rates, counts that occur over units of time , if X ~ Poisson (λt) is the expected count per unit of time and t is the monitoring time.
ppois(10,lambda = 5*3)## [1] 0.1184644