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adelina 88 [10]

3 years ago

15

Suppose you have a random sample of 50 of these animals. What is the probability that the mean life span of the sample group is

between 15 and 20 years?

1 answer:

katrin2010 [14]3 years ago

7 0

In what I know of this problem and what you told me I think there is a 1/2 chance!

think about it, if the life span is 15 years old with standard deviation of 3 then that means we can have bats from 12~18 years old and the mean of the 50 numbers can be anything between that depending on the results we got. Then half of values in interval 12~18 is over 15 which gives it a 1/2 probability!

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Fiona’s school has three hallways that make up three sides of a triangular building design. In the morning, Fiona walks 90 yards

lilavasa [31]

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Let a1, a2, a3, ... be a sequence of positive integers in arithmetic progression with common difference

Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

$a_2 = a_1 + 2$

$a_3 = a_2 + 2 = a_1 + 2\cdot2$

$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

$b_2 = 2b_1$

$b_3=2b_2 = 2^2 b_1$

$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

so the left side is

$2(a_1+a_2+\cdots+a_n) = 2c n + 2n(n-1) = 2n^2 + 2(c-1)n$

Also recall that

$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

so that the right side is

$b_1 + b_2 + \cdots + b_n = \displaystyle \sum_{k=1}^n 2^{k-1}b_1 = c(2^n-1)$

Solve for $c$ .

$2n^2 + 2(c-1)n = c(2^n-1) \implies c = \dfrac{2n^2 - 2n}{2^n - 2n - 1} = \dfrac{2n(n-1)}{2^n - 2n - 1}$

Now, the numerator increases more slowly than the denominator, since

$\dfrac{d}{dn}(2n(n-1)) = 4n - 2$

$\dfrac{d}{dn} (2^n-2n-1) = \ln(2)\cdot2^n - 2$

and for $n\ge5$ ,

$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

This means we only need to check if the claim is true for any $n\in\{1,2,3,4\}$ .

$n=1$ doesn't work, since that makes $c=0$ .

If $n=2$ , then

$c = \dfrac{4}{2^2 - 4 - 1} = \dfrac4{-1} = -4 < 0$

If $n=3$ , then

$c = \dfrac{12}{2^3 - 6 - 1} = 12$

If $n=4$ , then

$c = \dfrac{24}{2^4 - 8 - 1} = \dfrac{24}7 \not\in\Bbb N$

There is only one value for which the claim is true, $c=12$ .

3 0

2 years ago