下面的茎叶图显示了20名学生在一次满分为50分的班级测试中所获得的分数。 写出中位数分数。 答案 ... 求标准差。 答案 ... 如果获得优秀的标准是分数至少达到80%，求班级中获得优秀的学生百分比。 答案 ... % 贝尔有六枚10美分硬币和两枚50美分硬币。她从钱包里随机摸出2枚硬币，一次摸一枚。 完成答题区域中所示的概率树状图，答案用最简分数表示。 求这两枚硬币总值为60美分的概率。 答案 ... 如果再取出第三枚硬币，计算总金额为30美分的概率。 答案 ...

1. Identify the middle two values and find their average

There are $20$ pupils, so the median is the average of the $10$th and $11$th marks.

From the ordered stem-and-leaf diagram, the $10$th and $11$th marks are both $35$.

$\text{Median}=\frac{35+35}{2}=35$

1. Find the total and the mean of the 20 marks

The marks are
$10,12,23,23,24,27,27,32,35,35,35,36,37,39,41,46,47,48,48,50$.

$\sum x=675,\quad n=20$

$\bar{x}=\frac{675}{20}=33.75$

2. Find the variance using the standard deviation formula

$\sum x^2=25335$

Using
$\sigma^2=\frac{\sum x^2}{n}-\bar{x}^2,$

$\sigma^2=\frac{25335}{20}-(33.75)^2$

$=1266.75-1139.0625$

$=127.6875$

3. Take the square root of the variance

$\sigma=\sqrt{127.6875}$

$=11.3008\ldots$

So the standard deviation is $11.3$ to 3 significant figures.

1. Find the minimum mark needed for distinction

Distinction is at least $80\%$ of $50$ marks.

$80\% \times 50 = 0.8 \times 50 = 40$

So pupils who scored $40$ marks or more get distinction.

2. Count the pupils who scored at least 40 and express as a percentage

Marks at least $40$ are:

$41,\ 46,\ 47,\ 48,\ 48,\ 50$

Number of pupils with distinction $=6$.

$\frac{6}{20}\times 100\% = 30\%$

1. Find the probabilities for the first coin drawn

There are $6+2=8$ coins altogether.

$P(10\text{c})=\frac68=\frac34$

$P(50\text{c})=\frac28=\frac14$

2. Find the second-draw probabilities after a 10-cent coin is drawn first

If the first coin is $10\text{c}$, the remaining coins are $5$ ten-cent coins and $2$ fifty-cent coins, out of $7$ coins.

$P(10\text{c second}\mid 10\text{c first})=\frac57$

$P(50\text{c second}\mid 10\text{c first})=\frac27$

3. Find the second-draw probabilities after a 50-cent coin is drawn first

If the first coin is $50\text{c}$, the remaining coins are $6$ ten-cent coins and $1$ fifty-cent coin, out of $7$ coins.

$P(10\text{c second}\mid 50\text{c first})=\frac67$

$P(50\text{c second}\mid 50\text{c first})=\frac17$

1. Identify the outcomes that give a total of 60 cents

A total of $60$ cents is obtained from:

$10\text{c}$ then $50\text{c}$, or $50\text{c}$ then $10\text{c}$.

2. Multiply along each branch and add the two probabilities

$P(60\text{c})=\frac{3}{4}\times\frac{2}{7}+\frac{1}{4}\times\frac{6}{7}$

$=\frac{6}{28}+\frac{6}{28}$

$=\frac{12}{28}=\frac{3}{7}$

1. Identify the only coin combination that gives 30 cents

To get a total of $30$ cents from $3$ coins, the coins must be

$10\text{c},\ 10\text{c},\ 10\text{c}.$

2. Multiply the probabilities of drawing three 10-cent coins without replacement

There are initially $6$ ten-cent coins out of $8$ coins.

$P(10\text{c},10\text{c},10\text{c})=\frac{6}{8}\times\frac{5}{7}\times\frac{4}{6}$

$=\frac{5}{14}.$