Exercises: Continuous random variables

Parametric continuous distribution

Exercise 1 (The normal table) Let \(Z \sim N(0,1)\) be a standard normal random variable, and compute;

  1. \(P(Z<1.64)\)
  2. \(P(Z>-1.64)\)
  3. \(P(-1.96<Z)\)
  4. \(P(Z<2.36)\)
  5. An \(a\) such that \(P(Z<a) = 0.95\)
  6. A \(b\) such that \(P(Z>b) = 0.975\)

Note, this exercise can be solved using the standard normal table or using the R functions pnormand qnorm.

  1. \(P(Z<1.64) = P(Z \leq 1.64) = [\textrm{from table}] = 0.9495\)
  2. \(P(Z>-1.64) = [\textrm{symmetry}] = P(Z<1.64) = 0.9495\)
  3. \(P(-1.96<Z) = [\textrm{symmetry}] = P(Z<1.96) = [\textrm{from table}] = 0.975\)
  4. \(P(Z<2.36) = [\textrm{from table}] = 0.9909\)
  5. Standard normal table gives that \(a=1.64\)
  6. \(P(Z>b) = 0.975\), symmetry gives that \(P(Z<-b)=0.975\). Look-up in the standard normal table gives that \(-b=1.96\), hence \(b=-1.96\).

Exercise 2 (Exercise in standardization/transformation) If \(X \sim N(3,2)\), compute the probabilities

  1. \(P(X<5)\)
  2. \(P(3<X<5)\)
  3. \(P(X \geq 7)\)
  1. \(P(X<5) = P\left(\frac{X-3}{2} < \frac{5-3}{2}\right) = P(Z<1) = [\textrm{from table}] = 0.8413\)
  2. \(P(3<X<5) = P\left(\frac{3-3}{2} < \frac{X-3}{2} < \frac{5-3}{2}\right) = P(0<Z<1) = P(Z<1) - P(Z<0) = 0.8413 - 0.5000 = 0.3413\)
  3. \(P(X \geq 7) = P\left(\frac{X-3}{2} \geq \frac{7-3}{2}\right) = P(Z \geq 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228\)

Exercise 3 (Hemoglobin) The hemoglobin (Hb) value in a male population is normally distributed with mean 188 g/L and standard deviation 14 g/L.

  1. Men with Hb below 158 g/L are considered anemic. What is the probability of a random man being anemic?
  2. When randomly selecting 10 men from the population, what is the probability that none of them are anemic?
  1. \(P(Hb<158) = P(Z<\frac{158-188}{14}) = P(Z<-2.14) = [table] = 0.016\) or use R
pnorm(158, mean=188, sd=14)
[1] 0.01606229
  1. Probability of one man not being anemic; \(1-0.016 = 0.984\). Probability of 10 selected men not being anemic (binomial distribution) \(0.984^{10} = 0.85\)

Exercise 4 (Pill) A drug company is producing a pill, with on average 12 mg of active substance. The amount of active substance is normally distributed with mean 12 mg and standard deviation 0.5 mg, if the production is without problems. Sometimes there is a problem with the production and the amount of active substance will be too high or too low, in which case the pill has to be discarded. What should the upper and lower critical values (limits for when a pill is acceptable) be in order not to discard more than 1/20 pills from a problem free production?

\(H_0: \mu=12\) \(H_1: \mu \neq 12\)

Search \(x_{low}\) and \(x_{up}\) such that \(P(x_{low} < X < x_{up}) = 0.05\)

From table we know that \(P(-1.96 < Z < 1.96) = 0.05\)

\(Z = \frac{X - \mu_0}{\sigma_0} = \frac{X-12}{0.5}\), hence

\(-1.96 < Z < 1.96 \iff -1.96 < \frac{X-12}{0.5} < 1.96 \iff 12-1.96*0.5 < X < 12+1.96*0.5 \iff 11.02 < X < 12.98\)

The lower and upper critical values of active substance should be 11.02 and 12.98 mg.

Random sample

Exercise 5 (Exercise in distribution of sample mean) The total cholesterol in population (mg/dL) is normally distributed with \(\mu = 202\) and \(\sigma = 40\).

  1. How is the sample mean of a sample of 4 persons distributed?
  2. What is the probability to see a sample mean of 260 mg/dL or higher?
  3. Is there reason to believe that the four persons with mean 260 mg/dL were sampled from another population with higher population mean?
  1. \[\bar X = \frac{1}{4}\sum_{i=1}^4 X_i \\ X_i \sim N(\mu, \sigma) \\ \bar X \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) = N(202, 20) \]
  2. 0.0019
  3. Yes

Exercise 6 (Amount of active substance) The amount of active substance in a pill is stated by the manufacturer to be normally distributed with mean 12 mg and standard deviation 0.5 mg. You take a sample of five pills and measure the amount of active substance to; 13.0, 12.3, 12.6, 12.5, 12.7 mg.

[Note: a-c were already computed in the descriptive statistics session.]

  1. Compute the sample mean
  2. Compute the sample variance
  3. Compute the sample standard deviation
  4. compute the standard error of mean, \(SEM\).
  5. If the manufacturers claim is correct, what is the probability to see a sample mean as high as in a) or higher?
  1. 12.62
  2. 0.067
  3. 0.26
  4. 0.22
  5. 0.0028
  6. Sample mean
x <- c(13.0, 12.3, 12.6, 12.5, 12.7)
n <- length(x)
(m <- sum(x)/n)
[1] 12.62
  1. Sample variance
[1] 0.067
  1. Sample standard deviation
[1] 0.2588436
  1. Standard error of mean, SEM
[1] 0.2236068

Note, here the known standard deviation, \(\sigma=0.5\) is used.