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3 years ago

7

A bag contains 8 purple beads 2 blue beads and 2 pink beads. The probability of randomly drawing a pink beads is 1/6 . What is t

he probability of not drawing a pink bead

2 answers:

Dafna11 [192]3 years ago

7 0

8+2+2=12
Pink=2
2/12=1/6
10/12=5/6
The probability of not drawing a pink bead is 5/6.

otez555 [7]3 years ago

4 0

The probability of. Or picking pink would be 1 - the probability of picking pink:

1 - 1/6 = 5/6

Answer: 5/6

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Answer:

$Median = 90.4$

$Q_1 = 88.6$

$Q_3 = 92.2$

Step-by-step explanation:

Given

The above data

Required

- A stem and leaf display

- The median

- The quartiles

First, determine the range of the data

$Smallest = 83.4$

$Highest = 100.3$

Next, group each dataset base on common whole numbers.

So, we have:

$83.4$

$84.3\ 84.3$

$85.3$

$86.7\ 86.7\ 86.7$

$87.4\ 87.5\ 87.6\ 87.7\ 87.8\ 87.9$

$88.2\ 88.3\ 88.3\ 88.3\ 88.4\ 88.5\ 88.5\ 88.6\ 88.6\ 88.7\ 88.9$

$89.0\ 89.2\ 89.3\ 89.3\ 89.6\ 89.7\ 89.8\ 89.8\ 89.9\ 89.9$

$90.0\ 90.1\ 90.1\ 90.1\ 90.3\ 90.4\ 90.4\ 90.4\ 90.5\ 90.6\ 90.7\ 90.8\ 90.9$

$91.0\ 91.0\ 91.0\ 91.1\ 91.1\ 91.1\ 91.2\ 91.2\ 91.2\ \ 91.5\ 91.6\ 91.6\ 91.8\ 91.8$

$92.2\ 92.2\ 92.2\ 92.3\ 92.6\ 92.7\ 92.7\ 92.7$

$93.0\ 93.2\ 93.3\ 93.3\ 93.4\ 93.7$

$94.2\ 94.2\ 94.4\ 94.7$

$96.1\ 96.5$

$98.8$

$100.3$

Next, we construct the stem and leaf plot.

The whole numbers will be the stem while the decimal parts will be the leaf.

So, we have:

$\begin{array}{ccc}{Stem} & {} & {Leaf} & {83} & {|} & {.4} & {84} & {|} & {.3\ .3} & {85} & {|} & {.3} & {86} & {|} & {.7\ .7\ .7} & {87} &{|} & {.4\ .5\ .6\ .7\ .8\ .9} & {88} & {|} & {.2\ .3\ .3\ .3\ .4\ .5\ .5\ .6\ .6\ .7\ .9} &{89} & {|} & {.0\ .2\ .3\ .3\ .6\ .7\ .8\ .8\ .9\ .9} & {90} & {|} &{.0\ .1\ .1\ .1\ .3\ .4\ .4\ .4\ .5\ .6\ .7\ .8\ .9} & {91} &{|}&{.0\ .0\ .0\ .1\ .1\ .1\ .2\ .2\ .2\ .5\ .6\ .6\ .8\ .8} &{92} &{|} &{.2\ .2\ .2\ .3\ .6\ .7\ .7\ .7} \ \end{array}$

$\begin{array}{ccc} {93} & {|} & {.0\ .2\ .3\ .3\ .4\ .7} & {94} &{|} & {.2\ .2\ .4\ .7} &{96} & {|} & {.1\ .5} & {98} & {|} & {.8} & {100} &{|} &{.3} \ \end{array}$

From the above plot,

$n = 83$

The median is calculated as:

$Median = \frac{n+1}{2}th$

$Median = \frac{83+1}{2}th$

$Median = \frac{84}{2}th$

$Median = 42nd$

i.e. the median is at the 42nd position.

From the above stem and leaf plot.

The 42nd position is at stem 90 and the leaf .4

So the median is:

$Median = 90.4$

The lower quartile (Q) is calculated as:

$Q_1 = \frac{n+1}{4}th$

$Q_1 = \frac{83+1}{4}th$

$Q_1 = \frac{84}{4}th$

$Q_1 = 21st$

i.e. the lower quartile is at the 21st position.

From the above stem and leaf plot.

The 42nd position is at stem 88 and the leaf .6

So the lower quartile is:

$Q_1 = 88.6$

The upper quartile (Q3) is calculated as:

$Q_3 = 3 * \frac{n+1}{4}th$

$Q_3 = 3 * \frac{83+1}{4}th$

$Q_3 = 3 * \frac{84}{4}th$

$Q_3 = 3 * 21th$

$Q_3 = 63rd$

i.e. the upper quartile is at the 63rd position.

From the above stem and leaf plot.

The 63rd position is at stem 92 and the leaf .2

So the upper quartile is:

$Q_3 = 92.2$

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Answer:

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3 years ago

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