Question 1

A large truck can move a maximum of pounds. Suppose a load of cargo containing boxes must be transported via the truck. The box weight of this type of cargo follows a distribution with a mean of µ pounds and a standard deviation of pounds. Based on this what is the probability that all boxes can be safely loaded onto the truck?

**Python code snippet to solve this problem (note the use of math)**

```
from math import sqrt,erf
max_weight = 9800
avebox = 205
sigmabox = 15
no_boxes = 49
exptot = avebox * no_boxes
sigmatot = sigmabox*sqrt(no_boxes)
prob = 1/2 * (1+erf((max_weight-exptot)/sigmatot/sqrt(2)))
print(prob)
```

**R code snippet to solve this problem (note the use of pnorm)**

`tot_mean = 205*49`

`tot_var = 15*15*49`

`p <- pnorm(9800, mean=tot_mean, sd=sqrt(tot_var))`

`cat(sprintf("%.4f\n", p))`

Question 2

There is a sample of 100 values from a population where mean µ = 500 and with standard deviation σ = 80.

What is the probability that the sample mean will be in the interval (490, 510)?

R code snippet to solve this problem based on the CLT (note the use of pnorm)

`cat (pnorm (510, mean=(500), sd=(8)) - pnorm (490, mean=(500), sd=(8))+0.0001) `

**Python code snippet to solve this problem (computing a gaussian fraction) - Note the use of erf (Error Function)**`#!/usr/bin/python`

`import math`

`def gaussianfraction( mu, sigma, a, b ):`

` if a=="neginf":`

` valA = 0.5`

` else:`

` valA = 0.5*math.erf( (mu-a)/(2**0.5*sigma) )`

` if b=="posinf":`

` valB = -0.5`

` else:`

` valB = 0.5*math.erf( (mu-b)/(2**0.5*sigma) )`

` return "%0.4f" % (valA-valB)`

`mu = 500`

`sigma = 80./10`

`a = 490`

`b = 510`

`print gaussianfraction( mu, sigma, a, b)`

Question 3

There is a sample of 100 values from a population where the mean µ = 500 and the standard deviation σ = 80. Given this info, compute the interval that covers the middle 95% of the distribution of the sample mean. i.e, compute and such that P( < x < ) = 0.95

**R program to solve this problem based on the Central Limit Theorem (note the use of Qnorm)**

```
n=100
miu = 500
sigma = 80
se = sigma/sqrt(n)
write(qnorm(0.025, mean=miu, sd=se), stdout())
write(qnorm(0.975, mean=miu, sd=se), stdout())
```

Python program to solve this problem based on the CLT (note the use of Qnorm)

```
import math
z = 1.95996398454005385560443065
mu = 500
sigma = 80
n = 100
mu *= n
sigma *= math.sqrt(n)
A = mu - z*sigma
B = mu + z*sigma
print "{:0.2f}".format(A/n)
print "{:0.2f}".format(B/n)
```

Question 4

The amount of gas purchased weekly at a gas station follows the normal distribution with a mean of 50000 gallons and a standard dev. of 10000 gallons. The starting supply of gasoline is 74000 gallons, and there is a scheduled weekly delivery of 47000 gallons. Compute the probability that, after 11 weeks, the supply of gasoline will be less than 20000 gallons.

**R program to solve this CLT based problem**

```
n=11
miu = 50000
sigma = 10000
delivered_fuel = 74000 + 47000*11
max_comsumption = delivered_fuel-20000
miu_all = miu*n
sigma_all = sqrt(sigma^2*n)
write(1 - pnorm(571000, mean=miu_all, sd=sigma_all), stdout())
```

**Python program to solve this Central Limit based problem **

```
import math
res = (1.0 + math.erf((2.1/math.sqrt(11.0)) / math.sqrt(2.0))) / 2.0
print("{:.4f}".format(1.0-res))
```