What is the variance of the distribution of the average an IID draw of n observations from a population with mean \({\mu}\) and variance \({\sigma^2}\) ?
Answer : \({\frac{\sigma^2}{n}}\).
Suppose that diastolic blood pressures (DBPs) from men aged 35-44 are normally distributed with a mean of 80mmHg and a standard deviation of 10 mmHg. About what is the probability that a random 35-44 year old has a DBP less than 70 ?
Answer : We need to calculate \({P(X \leq 70)}\), knowing that \({X \sim \mathcal{N}(80,100)}\).
pnorm(70,mean=80,sd=10,lower.tail = TRUE)## [1] 0.1586553The searched probability is around 16%.
Brain 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 ?
Answer : We just need to calculate the quantile corresponding to a probability of 0.95, knowing that the brain volume follows a normal law \({\mathcal{N}(1100,{75^2})}\).
qnorm(0.95,mean=1100,sd=75,lower.tail = TRUE)## [1] 1223.364So the \({95^{th}}\) quantile for this distribution is approximately 1123.
Refer to the previous question. Brain volume for adult women is normally distributed with a mean of 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 th 95th percentile of the distribution of the sample mean ?
Answer : As the number of people is large enough, we can consider that the sample mean follows a normal distribution \({\mathcal{N}(\mu,{\sigma^2}/n)}\), where \({\mu = 1100}\), \({\sigma = 75}\) and \({n=100}\).
qnorm(0.95,mean=1100,sd=75/10,lower.tail = TRUE)## [1] 1112.336So the \({95^{th}}\) quantile for this distribution mean is approximately 1113.
You flip a fair coin 5 times. About what is the probability of getting 4 or 5 heads ?
Answer : As the coin is fair, the probability of getting 1 head at each flip is 0.5. The probability of getting at least 4 heads after 5 flips canbe computed using the binomial law :
\({p = \mathcal{C}_{5}^{4}.(0.5)^4(1-0.5) + \mathcal{C}_{5}^{5}.(0.5)^5}\)
pbinom(3,size=5,prob=0.5, lower.tail = FALSE)## [1] 0.1875So the probability to get at least 4 heads after 5 flips with a fair coin is approximately 19%.
The 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.
Answer : The standard error of the mean is \({\frac{\sigma}{\sqrt{n}}}\), where \({\sigma = 10}\) and \({n = 100}\). So the value is 1. We then want to measure the probability of the RDI being inside 1 standard deviation around the mean. Thus it should be around 68%.
Consider a standard uniform density. The mean for this density is 0.5 and the variance is 1/12. You sample 1000 observations from this distribution and take the sample mean, what value would you expect it to be near ?
Answer : Using the LLN, the value should be approximately 0.5.
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 of fewer people ?
ppois(10, lambda=15)## [1] 0.1184644Answer : The probability is approximately 12%.