The change in depth of a river from one day to the next, measured in cm at a specific location, is a continuous uniform random variable $X$ with the following density function: $f(x)=k$ for $-2 \le x \le 2$ and $f(x)=0$ elsewhere.
What is the value of k?
We are given the density function $f(x) = k$ if $-2 \leq k \leq 2$ and 0 otherwise. In order to find the value of $k$, we must keep in mind that the function must integrate to 1, c.f. the definition of a continuous density function, Chapter 2. We then get:
The change in depth of a river from one day to the next, measured in cm at a specific location, is a continuous uniform random variable $X$ with the following density function: $f(x)=k$ for $-2 \le x \le 2$ and $f(x)=0$ elsewhere.
What is $P(X > 1)$:
We are given the density function $f(x) = k$ if $-2 \leq k \leq 2$ and 0 otherwise. So we have $X \sim Uniform(-2,2)$. We calculate:
\(P(X > 1) = \frac{1}{4} \text{ From geometrical considerations, or}\)
\(P(X > 1) = \int_1^2 \frac{1}{4} dx = \frac{1}{4}\)
Correct answer is 2.
Question 3 of 4
In a consumer survey performed by a newspaper, 20 different groceries (products) were purchased in a grocery store. Discrepancies between the price appearing on the sales slip and the shelf price were found in 6 of these purchased products.
At the same time a customer buys 3 random (different) products within the group consisting of the 20 goods in the store. The probability that no discrepancies occurs for this customer is:
Let $ X $ denote the number of discrepancies when purchasing 3 random (different) products within the group of the 20 goods in the store. $X$ then follows the hypergeometric distribution (NOT the binomial!!) (Why not binomial: Because you don’t potentially by two goods of the same kind - you DO buy 3 DIFFERENT ones and hence having bought one - you do NOT “put it back” again and the randomly select - it is WITHOUT replacement). So:
\(P(X=0)=\frac{\left(\begin{array}{c}14\\3\end{array}\right)\left(\begin{array}{c}6\\0\end{array}\right)}{\left(\begin{array}{c}20\\3\end{array}\right)}
=\frac{14\cdot 13\cdot 12\cdot 3\cdot 2}
{20\cdot 19\cdot 18\cdot 3\cdot 2}=\frac{91}{15\cdot 19}=0.3192982\)
Hence the answer is 5:
0.319
Question 4 of 4
A package of chocolate consists of 8 chocolate bars. It is assumed that the weight of the individual bars follows a normal distribution with mean weight of $\mu=100$ g and standard deviation $\sigma=1$ g.
Below are figures of four distributions: (a, b, c, and d, from top left to bottom right)
Which of the four figures is showing the distribution of the weight of the chocolate packages?
We need to find the distribution of the sum of eight random variables, where each follows the normal distribution with mean 100 and variance 1. Using the rules of means and variances of Chapter2, and assuming that the weights of the pieces of chocolate are independent we find
Also, the sum of normally distributed variables is itself normally distributed. Thus $Y \sim N(800, 8)$. The standard deviation of $Y$ is $\sqrt{8} = 2\sqrt{2} \approx 2.83$. 2.5% of the probability mass lies to each side of the interval $[800 \pm 1.96 \cdot 2.83]$. Hence the correct distribution is the symmetric distribution which has almost all its mass between 794.5 and 805.5, but still some (2.5% on each side) mass outside of that interval.
