Study questions platform-wide or filter by specific tests with correct answers revealed.
Five observations have mean $= 4$ and variance $= 5.2$. Three of the observations are $3, 4, 4$. Find the absolute difference between the remaining two observations.
Step 1 โ Find the sum: $\sum x_i = 5\times4=20$. Known three: $3+4+4=11$. So $x_4+x_5=9$.
Step 2 โ Find sum of squares: Variance $=\frac{\sum x_i^2}{n}-\bar{x}^2 \Rightarrow \frac{\sum x_i^2}{5}-16=5.2 \Rightarrow \sum x_i^2=106$.
Known: $3^2+4^2+4^2=9+16+16=41$. So $x_4^2+x_5^2=65$.
Step 3 โ Solve: $x_4+x_5=9$ and $x_4^2+x_5^2=65$.
$(x_4+x_5)^2=81 \Rightarrow 2x_4x_5=81-65=16 \Rightarrow x_4x_5=8$.
$(x_4-x_5)^2=(x_4+x_5)^2-4x_4x_5=81-32=49$
$|x_4-x_5|=\mathbf{7}$
Two cards are drawn one after another with replacement from a well-shuffled standard deck of 52 cards. Let $X$ be the number of aces obtained. Find $P(X=1)+P(X=2)$.
$p=P(\text{ace})=\dfrac{4}{52}=\dfrac{1}{13}$, $q=\dfrac{12}{13}$. Binomial with $n=2$.
$P(X=1)=\binom{2}{1}\cdot\dfrac{1}{13}\cdot\dfrac{12}{13}=\dfrac{24}{169}$
$P(X=2)=\binom{2}{2}\cdot\left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$
$P(X=1)+P(X=2)=\dfrac{24}{169}+\dfrac{1}{169}=\dfrac{25}{169}$
Sign in to join the conversation and share your thoughts.
Log In to Comment