Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 3.3 AMC 10/12 Find the answer

French mathematician Poincaré is a person who likes to eat bread. He goes to the same bakery every day to buy a loaf of bread. The baker at the bakery claims that the average weight of the bread he sells is $1000g$ , with a fluctuation of no more than $50g$ . In mathematical terms, this statement can be expressed as: the weight of each loaf of bread follows a normal distribution with an expectation of $1000g$ and a standard deviation of $50g.

$(1)$ Given the following conclusion: If $X\sim N(\mu, \sigma^2)$ , randomly select $k$ data points from the values of $X$ ( $k\in \mathbb{N}^*, k\geq 2$ ), and denote the average of these $k$ data points as $Y$ , then the random variable $Y$ follows $N(\mu, \frac{{\sigma^2}}{k})$ . Use this conclusion to solve the following problems:

$(i)$ Assuming the baker's claim is true, randomly purchase $25$ loaves of bread. Let the average weight of these $25$ loaves be $Y$ , find $P(Y\leq 980)$ ;

$(ii)$ Poincaré weighs and records the bread he buys every day. After $25$ days, all the data fall within $(950, 1050)$ , and the calculated average weight of the $25$ loaves is $978.72g$ . Poincaré reported the baker based on this data. Explain from a probability perspective why Poincaré reported the baker;

$(2)$ Assuming there are two identical boxes containing bread (except for the color, everything else is the same), it is known that the first box contains a total of $6$ loaves of bread, with $2$ black loaves; the second box contains a total of $8 loaves of bread, with$ 3 $black loaves. Now, randomly select a box, and then randomly pick$ 2$ loaves of bread from that box. Find the distribution table of the number of black loaves drawn and the mathematical expectation.

Given:
$(1)$ If a random variable $\eta$ follows a normal distribution $N(\mu, \sigma^2)$ , then $P(\mu -\sigma \leq \eta \leq \mu +\sigma) = 0.8627$ , $P(\mu -2\sigma \leq \eta \leq \mu +2\sigma) = 0.9545$ , $P(\mu -3\sigma \leq \eta \leq \mu +3\sigma) = 0.9973$ ;
$(2)$ Events with a probability less than $0.05$ are usually referred to as small probability events, which are unlikely to occur.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

### Solution:

#### Part 1(i)

Given that the weight of each loaf of bread follows a normal distribution with an expectation of $1000g$ and a standard deviation of $50g$ , and we are considering the average weight of $25$ loaves of bread:

1. Calculate the variance of the average weight $Y$ :
$\frac{50^2}{25} = 100$

2. Therefore, the distribution of $Y$ is:
$Y \sim N(1000, 10^2)$

3. Given the property of the normal distribution:
$P(\mu -2\sigma \leq \eta \leq \mu +2\sigma) = 0.9545$

4. The probability that $\eta$ is less than $\mu - 2\sigma$ is:
$P(\eta \leq \mu -2\sigma) = \frac{1-0.9545}{2} = 0.02275$

5. Since $980 = 1000 - 2 \times 10$ , we find:
$P(Y\leq 980) = P(Y\leq \mu -2\sigma) = 0.02275$

Therefore, the probability that the average weight of these $25$ loaves is less than or equal to $980g$ is $\boxed{0.02275}$ .

#### Part 1(ii)

Given that Poincaré calculated the average weight of $25$ loaves of bread as $978.72g$ , which is less than $980g$ , and considering the probability found in part (i):

1. The probability that the average weight is less than or equal to $980g$ is $0.02275$ .

2. Since $0.02275 < 0.05$ , this event is considered a small probability event, which is unlikely to occur under normal circumstances.

Therefore, from a probability perspective, Poincaré reported the baker because the observed average weight being significantly less than claimed is a rare event, indicating that the baker's claim might not be true.

#### Part 2

Considering two boxes, one with $6$ loaves of bread (including $2$ black loaves) and the other with $8$ loaves (including $3$ black loaves), and randomly selecting $2$ loaves of bread from a randomly chosen box:

1. The probability of drawing $0$ black loaves is calculated as:
$P(\xi = 0) = \frac{1}{2} \times \frac{4}{6} \times \frac{3}{5} + \frac{1}{2} \times \frac{5}{8} \times \frac{4}{7} = \frac{53}{140}$

2. The probability of drawing $1$ black loaf is:
$P(\xi = 1) = \frac{1}{2} \times \frac{2}{6} \times \frac{4}{5} \times 2 + \frac{1}{2} \times \frac{3}{8} \times \frac{5}{7} \times 2 = \frac{449}{840}$

3. The probability of drawing $2$ black loaves is:
$P(\xi = 2) = \frac{1}{2} \times \frac{2}{6} \times \frac{1}{5} + \frac{1}{2} \times \frac{3}{8} \times \frac{2}{7} = \frac{73}{840}$

4. The distribution table of $\xi$ is thus:

$\xi$	$0$	$1$	$2$
$P$	$\frac{53}{140}$	$\frac{449}{840}$	$\frac{73}{840}$

5. The mathematical expectation of

\xi

is calculated as:

E(\xi) = 0 \times \frac{53}{140} + 1 \times \frac{449}{840} + 2 \times \frac{73}{840} = \frac{17}{24}

Therefore, the mathematical expectation of the number of black loaves drawn is

\boxed{\frac{17}{24}}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.