Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, x y, be 8 and 16 respectively.

Q: Why can't we just use the definition of variance $\frac{\sum (x_i - \bar{x})^2}{n}$? It seems more intuitive.

It is intuitive, but computationally heavy. Expanding $(x_i - 8)^2$ for seven terms is a recipe for a calculation error. The computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ is the professional's choice for a reason—it isolates the sum of squares, which is exactly what we need for our system of equations.

Q: In the probability part, why does 'without replacement' not change the combination formula $\binom{6}{2}$?

Great observation! Whether you pick one, then another, or grab two at once, the set of two numbers you end up with is the same. Order doesn't matter for the set, so $\binom{n}{r}$ is perfectly valid. The 'without replacement' condition simply ensures we don't pick the same number twice, which is naturally handled by the combination formula.

Q: Is the complement method always better?

Not always, but here, it is a lifesaver. Counting cases where the smaller number is 1, 2, or 3 requires multiple sub-cases. Counting cases where the smaller number is 4, 5, or 6 is just one simple scenario. Always look for the path of least resistance in probability problems.

Question

Let the mean and variance of 7 observations $2, 4, 10, x, 12, 14, y, x > y$, be 8 and 16 respectively. Two numbers are chosen from $\{1, 2, 3, x - 4, y, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :

Options:

(A) $\frac{3}{5}$

(B) $\frac{1}{3}$

(C) $\frac{2}{5}$

(D) $\frac{4}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a problem that is not just a calculation, but a beautiful dance between two pillars of mathematics: Statistics and Probability. We are given a set of seven observations: $2, 4, 10, x, 12, 14, y$.

We know the mean is $8$ and the variance is $16$. Our mission is to unmask $x$ and $y$ and then navigate a probability challenge.

### The Detective Work

We begin by treating the mean as our first clue. The mean is the arithmetic average, the balance point of our data:

$$
\frac{2 + 4 + 10 + x + 12 + 14 + y}{7} = 8
$$

Summing the known constants gives us $42$. Thus, $42 + x + y = 56$, which simplifies to our first elegant equation:

$$
x + y = 14
$$

Now, we turn to the variance. We use the computational formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$. Substituting our known values:

$$
\frac{2^2 + 4^2 + 10^2 + x^2 + 12^2 + 14^2 + y^2}{7} - 8^2 = 16
$$

This simplifies to:

$$
\frac{460 + x^2 + y^2}{7} = 80
$$

Multiplying by $7$ and subtracting $460$, we arrive at:

$$
x^2 + y^2 = 100
$$

### The Algebraic Elegance

We have a system: $x + y = 14$ and $x^2 + y^2 = 100$. Let's use the identity $(x+y)^2 = x^2 + y^2 + 2xy$.

Plugging in our values:

$$
14^2 = 100 + 2xy \Rightarrow 196 = 100 + 2xy \Rightarrow xy = 48
$$

We need two numbers that sum to $14$ and multiply to $48$. The factors of $48$ are $6$ and $8$. Since the problem dictates $x > y$, we assign **$x = 8$** and **$y = 6$**.

### The Probability Challenge

With $x=8$ and $y=6$, our set $S = \{1, 2, 3, x-4, y, 5\}$ transforms into $S = \{1, 2, 3, 4, 6, 5\}$, which is simply the set of the first six natural numbers. We are choosing two numbers without replacement.

The total ways to choose two numbers is:

$$
\binom{6}{2} = 15
$$

We want the probability that the smaller number is less than $4$. The complement of 'smaller number $< 4$' is 'smaller number $\ge 4$'.

If the smaller number is at least $4$, then both numbers must be chosen from the set $\{4, 5, 6\}$. The number of ways to choose two from these three is:

$$
\binom{3}{2} = 3
$$

Therefore, the favorable outcomes are $15 - 3 = 12$. The final probability is:

$$
P = \frac{12}{15} = \mathbf{\frac{4}{5}}
$$

Select Answer:

Visualized Solution

Understanding the Observations

Formulating the Mean Equation

Setting up the Variance Equation

Simplifying for $x^{2} + y^{2}$

Solving for $x$ and $y$

Finding the Values of $x$ and $y$

Constructing the New Set

Evaluating the New Set

Calculating Total Outcomes

Favorable Outcomes using Complement

Calculating Favorable Ways

Final Probability Calculation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Detective Work

The Algebraic Elegance

The Probability Challenge

Similar Questions

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is :

Let the mean and variance of 8 numbers $x, y, 10, 12, 6, 12, 4, 8$ be 9 and 9.25 respectively. If $x > y$ , then $3 x - 2 y$ is equal to ______.

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is :

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :

Let the Mean and Variance of five observations $x_{1} = 1, x_{2} = 3, x_{3} = a, x_{4} = 7$ and $x_{5} = b, a > b$ , be 5 and 10 respectively. Then the Variance of the observations $n + x_{n}, n = 1, 2, \dots, 5$ is

Let the mean and the variance of $5$ observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be $\frac{24}{5}$ and $\frac{194}{25}$ respectively. If the mean and variance of the first $4$ observation are $\frac{7}{2}$ and $a$ respectively, then $(4 a + x_{5})$ is equal to:

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :

The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and $a$ and $b$ are respectively mean and variance of remaining 6 observation, then $a + 3 b - 5$ is equal to ______.

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to

If the mean and variance of eight numbers $3, 7, 9, 12, 13, 20, x$ and $y$ be 10 and 25 respectively then $x y$ is equal to ________.