Introduction to Statistical Thinking

Let \(X\) be the total weight of 8 people. By the assumption, \(X \sim \mbox{Normal}(560, 57^2)\text{.}\) We are interested in the probability \(\Prob(X > 650)\text{.}\) This probability is equal to the difference between 1 and the probability \(\Prob(X \leq 650)\text{.}\) We use the function “pnorm" in order to carry out the computation:

1 - pnorm(650,560,57)

## [1] 0.05717406

We get that the probability that the total weight of 8 people exceeds 650kg is equal to 0.05717406.

Let \(Y\) be the total weight of 9 people. By the assumption, \(Y \sim \mbox{Normal}(630, 61^2)\text{.}\) We are interested in the probability \(\Prob(Y > 650)\text{.}\) This probability is equal to the difference between 1 and the probability \(\Prob(Y \leq 650)\text{.}\) We use again the function “pnorm" in order to carry out the computation:

1 - pnorm(650,630,61)

## [1] 0.3715054

We get that the probability that the total weight of 9 people exceeds 650kg is much higher and is equal to 0.3715054.

Again, \(X \sim \mbox{Normal}(560, 57^2)\text{,}\) where \(X\) is the total weight of 8 people. In order to find the central region that contains 80% of the distribution we need to identify the 10%-percentile and the 90%-percentile of \(X\text{.}\) We use the function “qnorm" in the code:

qnorm(0.1,560,57)
qnorm(0.9,560,57)

## [1] 486.9516
## [1] 633.0484

The requested region is the interval (486.9516, 633.0484).

As before, \(Y \sim \mbox{Normal}(630, 61^2)\text{,}\) where \(Y\) is the total weight of 9 people. In order to find the central region that contains 80% of the distribution we need to identify the 10%-percentile and the 90%-percentile of \(Y\text{.}\) The computation this time produces:

The probability \(\Prob(X > 11)\) can be computed as the difference between 1 and the probability \(\Prob(X \leq 11)\text{.}\) The latter probability can be computed with the function “pbinom":

1 - pbinom(11,27,0.32)

## [1] 0.1203926

Therefore, \(\Prob(X > 11) = 0.1203926\text{.}\)

Refer again to the probability \(\Prob(X > 11)\text{.}\) A formal application of the Normal approximation replaces in the computation the Binomial distribution by the Normal distribution with the same mean and variance. Since \(\Expec(X) = n \cdot p = 27 \cdot 0.32 = 8.64\) and \(\Var(X) = n \cdot p \cdot (1-p) = 27 \cdot 0.32 \cdot 0.68 = 5.8752\text{.}\) If we take \(X \sim \mbox{Normal}(8.64,5.8752)\) and use the function “pnorm" we get:

1 - pnorm(11,27*0.32,sqrt(27*0.32*0.68))

## [1] 0.1651164

Therefore, the current Normal approximation proposes \(\Prob(X > 11) \approx 0.1651164\text{.}\)

The continuity correction, that consider interval of range 0.5 about each value, replace \(\Prob(X > 11)\text{,}\) that involves the values \(\{12, 13, \ldots, 27\}\text{,}\) by the event \(\Prob(X > 11.5)\text{.}\) The Normal approximation uses the Normal distribution with the same mean and variance. Since \(\Expec(X) = 8.64\) and \(\Var(X) = 5.8752\text{.}\) If we take \(X \sim \mbox{Normal}(8.64,5.8752)\) and use the function “pnorm" we get:

1 - pnorm(11.5,27*0.32,sqrt(27*0.32*0.68))

## [1] 0.1190149

The Normal approximation with continuity correction proposes \(\Prob(X > 11) \approx 0.1190149\text{.}\)

The Poisson approximation replaces the Binomial distribution by the Poisson distribution with the same expectation. The expectation is \(\Expec(X) = n \cdot p = 27 \cdot 0.32 = 8.64\text{.}\) If we take \(X \sim \mbox{Poisson}(8.64)\) and use the function “ppois" we get:

1 - ppois(11,27*0.32)

## [1] 0.1635232

Therefore, the Poisson approximation proposes \(\Prob(X > 11) \approx 0.1651164\text{.}\)