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eimsori [14]
3 years ago
14

There are 27 chocolates in a box, all identically shaped. There 4 are filled with nuts, 8 with caramel, and 15 are solid chocola

te. You randomly select one piece, eat it, and then select a second piece. Find the probability of selecting a solid chocolate candy followed by a nut candy.
Mathematics
1 answer:
AlekseyPX3 years ago
6 0

Answer:

Therefore the required probability is \frac{30}{351}.

Step-by-step explanation:

Two events are dependents event if the occurrence of one of them has effect on the probability of the other.

If A and B are dependents,

then,

P(AB)=P(A)P(B).

Given that,

There are 27 chocolates in a box.

Number of nuts chocolates = 4

Number of caramel chocolates = 8

Number of solid chocolates= 15.

The probability of that a solid candy is drawn is

=\frac{\textrm{Number of solid chocolate}}{\textrm{Total number of chocolate}}

=\frac{15}{27}.

After selecting a solid chocolate, the number of chocolate is= (27-1)=26.

The probability that a nut candy is drawn is

=\frac{\textrm{Number of nut chocolate}}{\textrm{Total number of chocolate}}

=\frac{4}{26}

=\frac{2}{13}

Therefore the required probability is

=\frac{15}{27}\times\frac{2}{13}

=\frac{30}{351}

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Answer:

(a) See below

(b) r = 0.9879  

(c) y = -12.629 + 0.0654x

(d) See below

(e) No.

Step-by-step explanation:

(a) Plot the data

I used Excel to plot your data and got the graph in Fig 1 below.

(b) Correlation coefficient

One formula for the correlation coefficient is  

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}

The calculation is not difficult, but it is tedious.

(i) Calculate the intermediate numbers

We can display them in a table.

<u>    x   </u>    <u>      y     </u>   <u>       xy     </u>    <u>              x²    </u>   <u>       y²    </u>

   36       0.22              7.92               1296           0.05

   67        0.62            42.21              4489           0.40

   93         1.00            93.00           20164           3.46

 433        11.8          5699.4          233289        139.24

 887      29.3         25989.1          786769       858.49

1785      82.0        146370          3186225      6724

2797     163.0         455911         7823209    26569

<u>3675 </u>  <u> 248.0  </u>    <u>   911400      </u>  <u>13505625</u>   <u> 61504        </u>

9965   537.81     1545776.75  25569715   95799.63

(ii) Calculate the correlation coefficient

r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}

(c) Regression line

The equation for the regression line is

y = a + bx where

a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy  - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} -  \dfrac{9\times 1545776.75  - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}

(d) Residuals

Insert the values of x into the regression equation to get the estimated values of y.

Then take the difference between the actual and estimated values to get the residuals.

<u>    x    </u>   <u>      y     </u>   <u>Estimated</u>   <u>Residual </u>

    36        0.22        -10                 10

    67        0.62          -8                  9

    93        1.00           -7                  8

   142        1.86           -3                  5

  433       11.8             19               -  7

  887     29.3             45               -16  

 1785     82.0            104              -22

2797    163.0            170               -  7

3675   248.0            228               20

(e) Suitability of regression line

A linear model would have the residuals scattered randomly above and below a horizontal line.

Instead, they appear to lie along a parabola (Fig. 2).

This suggests that linear regression is not a good model for the data.

4 0
3 years ago
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