Question

Question 1 The weight of a package delivered by a courier company is normally distributed with mean 15 kg and standard deviation 1.75 kg.

(a) There are 9% of the packages weigh heavier than w kg. Find the value of w. (4 marks)

(b) Suppose the shipping fee set by the courier company is $150 basic charge, together with an additional charge of $16 per kg of the package. Find the following summary statistics of the shipping fee: (I) Mean (II) Standard deviation (III) Variance (IV) Median (V) 91st percentile (6 marks)

Answer 1

a)

$z=\frac{x-\mu}{\sigma}$

$P(x>w)=1-P(x<w)=0.09$

$P(x<w)=1-0.09=0.91$

$z=1.34$

$P(x<w)=P(z<1.34)=\frac{w-15}{1.75}=0.91$

$w=0.91\cdot1.75+15=16.6$ kg

b)

X is cost of package

i)

mean:

$\mu_1=150+16\mu=150+16\cdot15=\$390$

iii)

Variance of weight:

$\sigma^2=\frac{\sum (x_i-\mu)^2}{n}=1.75^2$

Variance of fee:

$\sigma_1^2=\frac{\sum 16^2(x_i-\mu)^2}{n}=16^2\cdot1.75^2=784$

ii)

Standard deviation of fee:

$\sigma_1=\sqrt{784}=28$

iv)

The median is the middle point in a dataset—half of the data points are smaller than the median and half of the data points are larger.

for normal distribution:

median = mean = 389

v)

91st percentile is a score below which a 91% of scores in its frequency distribution falls

for normal distribution:

for $P=0.91$ :

$z=1.34$

$P(z<1.34)=0.91$

$z=\frac{X-\mu_1}{\sigma_1}=\frac{X-390}{28}=1.34$

91st percentile of fee:

$X_{91}=1.34\cdot28+390=427.52$