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Sav [38]
3 years ago
15

The probability of an outcome that lies within 68% of the mean is a good indicator that it lies in which standard deviation?

Mathematics
1 answer:
kotegsom [21]3 years ago
8 0
Yes the probability of an outcome that lies within 68% is a good indicator that lies in which standard deviation.

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5 0
3 years ago
Samples of rejuvenated mitochondria are mutated (defective) in 3% of cases. Suppose 15 samples are studied, and they can be cons
stiks02 [169]

Answer:

C) At most one sample is mutated

Step-by-step explanation:

If there are 15 samples, it means that 15 is the total (100%) of samples. Then, if we know that there is a chance that 3% are mutated, then we calculate the 3% of 15:

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5 0
3 years ago
In a group of 10 ​kittens, 5 are female. Two kittens are chosen at random. ​a) Create a probability model for the number of male
olya-2409 [2.1K]

Answer:

(b) Expected number of male kitten E(x) = 1

(c) Standard deviation is ±\frac{2}{3}.

Step-by-step explanation:

Given that,

Number of kittens in a group are 10. Out of these 5 are female.

To find:- (a) create a probability model of male kitten chosen.

    So,  total number of kitten = 10

           number of female kitten = 5    

then, Number of male kitten = 10-5= 5

Now total number of ways to choosing two kittens from group of 10 = ^{10} C_{2}

                                                                                                                 = \frac{10!}{2!\times8!}

                                                                                                                 = 45

for choosing (i) No male P(x=0) = P(Two female) =  \frac{^{5} C_{2}}{45}

                                                                           = \frac{2}{9}

                     (ii) 1 male P(x=1) = P(1 male and 1 female) = \frac{^{5} C_{1}\times^{5} C_{1}}{45} =  \frac{25}{45} =\frac{5}{9}

                     (iii) 2 male P(x=2)= P(2 male) =\frac{^{5} C_{2}}{45} =  \frac{2}{9}

      x_{i}                             0                     1                      2

P(X=x_{i})                           \frac{2}{9}                      \frac{5}{9}                      \frac{2}{9}

(b) what is expected number of male kittens chosen ?

    Expected number of male kitten E(x) =  o\times \frac{2}{9} + 1\times \frac{5}{9} + 2\times\frac{2}{9}

                                                                  = 1

(c) what is standard deviation ?

                              \sigma(x) = \sqrt{ (E(x^{2} )-(Ex)^{2})

No,                          

                               Ex^{2} =o\times \frac{2}{9} + 1\times \frac{5}{9} + 4\times\frac{2}{9}

                                       = \frac{13}{9}

                               \sigma(x)=\sqrt{\frac{13}{9} - 1 }=\sqrt{\frac{4}{9} }  

                                       = ± \frac{2}{3}

                               

6 0
3 years ago
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