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Gemiola [76]
3 years ago
11

What is the y-intercept of line 12?

Mathematics
2 answers:
Assoli18 [71]3 years ago
7 0

Answer:

y int. of line 1 = 0

y int. of line 2 = 4

Step-by-step explanation:

hope this helps! moo!

Pani-rosa [81]3 years ago
6 0

Answer:

Answers are below

Step-by-step explanation:

The y-intercept of line 1 is 0 because it crosses the y-axis at 0.

The y-intercept of line 2 is 4 because it crosses the y-axis at 4.

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Answer:

a) 108 people with a standard deviation of 3.286335

b) No

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d) See explanation below.

Step-by-step explanation:

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<em>\large P(k;n)=\binom{n}{k}p^kq^{n-k} </em>

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<em>\large \binom{n}{k}= combination of n elements taken k at a time. </em>

<em>p = probability that the event (“success”) occurs once </em>

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In this case, the event “success” is finding an American adult who had  chickenpox before adulthood with probability p=0.9

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The <em>standard deviation for this binomial distribution</em> is

\large s=\sqrt{npq}

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a)

We consider a random sample of 120 American adults. How many people in this sample would you expect to have had chickenpox in their childhood?

If about 90% of American adults had chickenpox before adulthood, we expect to find 90% of 120 = 0.9*120=108 people in the sample who had chickenpox in their childhood.

The  standard deviation would be

\large s=\sqrt{120*0.9*0.1=3.286335}

b)

Would you be surprised if there were 105 people who have had chickenpox in their childhood?

No, because 105 and 108 are in the interval [mean - s, mean +s]

c)

What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood?

The probability that 105 or fewer people in this sample have had chickenpox in their childhood is

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}

We compute this value easily with a spreadsheet and we get

\large \sum_{k=0}^{105}\binom{120}{k}0.9^k0.1^{120-k}=0.218163\approx 21.82\%

d)

How does this probability relate to your answer to part (b)?

A binomial distribution with np>5 and nq>5 with n the sample size as is the case here, behaves pretty much like a Normal distribution with mean np and standard deviation \large \sqrt{npq}, so around 60% of the data are in the interval  [mean -s, mean +s] and 40% outside, so roughly <em>20% of the data should be in [0, mean-s] </em>

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