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yuradex [85]
3 years ago
11

​34% of working mothers do not have enough money to cover their health insurance deductibles. You randomly select six working mo

thers and ask them whether they have enough money to cover their health insurance deductibles. The random variable represents the number of working mothers who do not have enough money to cover their health insurance deductibles.
Required:
Construct a binomial distribution using n= 0.6 and p=0.34
Mathematics
1 answer:
tamaranim1 [39]3 years ago
5 0

Answer:

solution below

Step-by-step explanation:

The question says 6 working mother's were selected so <u>n = 6</u><u> </u><u>not</u><u> </u><u>0</u><u>.</u><u>6</u>

We are expected to find

P(X = 0,1,2,3,4,4,6)

1. <u>When x = 0</u>

6C0*(0.34)⁰*(0.66)⁶

= 1 *1* 0.827

= 0.0827

<u>2. When X = 1</u>

6C1*(0.34)¹*(0.66)⁵

= 6 x 0.34 x 0.252

= 0.2555

<u>3. When X = 2</u>

6C2*(0.34)²*(0.66)⁴

= 15 x 0.1156 x 0.1897

= 0.3289

<u>4. When x = 3</u>

6C3*(0.34)³*(0.66)³

20 x 0.039304 x 0.2875

= 0.2599

<u>5. When X = 4</u>

6C4*(0.34)⁴*(0.66)²

= 15 x 0.01336 x 0.4356

= 0.8729

<u>6. When x = 5</u>

6C5*(0.34)⁵*(0.66)¹

= 6 x 0.0045 x 0.66

= 0.01782

<u>7. When x = 6</u>

6C6*(0.34)⁶*(0.66)⁰

1 x 0.0015 x 1

= 0.0015

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So, if you own 132 shares, you get 132*12.11=1598.52 dollars per year for all the stocks.

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Step-by-step explanation:

We know the following info from the problem

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We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

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(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

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And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

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