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ludmilkaskok [199]

3 years ago

10

There are 20 children in the cast of a class play, and 8 of the children are boys. of the boys, 4 have a speaking part in the pl

ay, and of the girls, 8 do not have a speaking part in the play. if a child from the cast of the play is chosen at random, what is the probability that the child has a speaking part?a. 2/5b. 1/2c. 3/5d. 3/4

2 answers:

S_A_V [24]3 years ago

7 0

Answer:

Step-by-step explanation:

kotykmax [81]3 years ago

5 0

<span>it all looks confusing when we try to juggle with all those numbers in the head. The problem can be solved systematically by constructing a contingency table.
</span>role/gender B G total
speaking...... 4 4 8
<span> silent............ 4 8 12
total............. 8 12 20
</span>Probability of a child having a speaking part is therefore
(4+4)/20=8/20=2/5
a. 2/5

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

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$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

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2 years ago

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maxonik [38]

Let f(x) = x² + 6x²-x+ 5 then ,

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According to the qn,

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R=0

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