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mote1985 [20]
4 years ago
5

A couple wants to have three children. Assume that the probabilities of a newborn being male or being female are the same and th

at the gender of one child does not influence the gender of another child. There are eight possible arrangements of girls and boys. What is the sample space for having three children (gender of the first, second, and third child)? All eight arrangements are (approximately) equally likely. The future parents are wondering how many boys they might get if they have three children. Give a probability model (sample space and probabilities of outcomes) for the number of boys. Follow the method of
Mathematics
1 answer:
liberstina [14]4 years ago
7 0

Answer:

Let the G represent the Girl child

and, B represents the Boy child.

Then Sample Space = {BGG, GBG, GGB, BBG, BGB, GBB, BBB, GGG}

Thus, There are 8 sample space.

Then, Probability that couple have 1 boy = {BGG, GBG, GGB}

=  3 ÷ 8 = 0.375

Probability that couple have 2 boys = { BBG, BGB, GBB}

= 3 ÷ 8 = 0.375

Probability that couple have 3 boys = {GGG}

= 1 ÷ 8 = 0.125

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Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let \mu = <u><em>population mean length of the board</em></u>.

So, Null Hypothesis, H_0 : \mu = 2564.0 millimeters    {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, H_A : \mu \neq 2564.0 millimeters      {means that the boards are either too long or too short}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

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where, \bar X = sample mean length of boards = 2559.5 millimeters

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             n = sample of boards = 26

So, <em><u>the test statistics</u></em> =  \frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }  ~   t_2_5

                                     =  -1.529    

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

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