Answer to Question #278674 in Statistics and Probability for Neo

Question

The early morning trading volumes (millions of shares) at Nairobi Securities Exchange for 18 days in January are as shown below. The probability distribution of trading volume is approximately normal. 216, 143, 221, 169, 198, 202, 210, 122, 187, 212, 232, 181, 199, 220, 258, 233, 219, 221. What is the probability that, on a randomly selected day, the early morning trading volume will be at most 189 million shares?

How many trading volumes should the company have to be in the top 35%?

Answer 1

Solution:

Given data: 216, 143, 221, 169, 198, 202, 210, 122, 187, 212, 232, 181, 199, 220, 258, 233, 219, 221

Count, N = 18

Sum, Σx = 3643

Mean, $\bar x=\mu=\dfrac{3643}{18}\approx202.39$

$\sigma=\sqrt{\dfrac{1}{N-1}\Sigma(x_i-\bar x)^2} \\=\sqrt{\dfrac{1}{18-1}\times(216-202.39)^2+...(221-202.39)^2} \\=\sqrt{1079.4281045752} \\\\\approx 32.85$

Variance, s²: 1079.4281045752

Now, $X\sim N(\mu,\sigma)$

(a):

$P(X \le 189)=P(z\le \dfrac{189-202.39}{32.85}) \\=P(z\le-0.41) \\=1-P(z\le 0.41) \\=1-0.65910 \\=0.3409$

(b):

$P(X>x)=0.35 \\\Rightarrow P(X\le x)=1-0.35=0.65 \\\Rightarrow P(z\le \dfrac{x-\mu}{\sigma})=0.65$

$\Rightarrow \dfrac{x-202.39}{32.85}=0.088$ [using z-score table]

$\Rightarrow x=205.2808$