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2 years ago

Each of the 6 cats in a pet store was weighed. Here are their weights (in pounds):

Mathematics

1 answer:

kobusy [5.1K]2 years ago

8 0

Answer:

Mean = 10.83

Median = 13

Step-By-Step Explanation:

Mean

(14, 5, 5, 13, 13, 15) ÷ 6 = 10.8333333

Shortened to 10.83

Median

The middle of 5, 5, 13, 13, 14, 15

(13+13)/2 = 13

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3 years ago

An example problem in a Statistics textbook asked to find the probability of dying when making a skydiving jump.

MArishka [77]

Answer:

(a) 0.999664

(b) 15052

Step-by-step explanation:

From the given data of recent years, there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths.

So, the probability of death is $\frac{21}{3000000}==0.000007$ .

Assuming, this probability holds true for each skydiving and does not change in the present time.

So, as every skydiving is an independent event having a fixed probability of dying and there are only two possibilities, the diver will either die or survive, so, all skydiving can be regarded as is Bernoulli's trial.

Denoting the probability of dying in a single jump by $q$ .

$q=7\times 10^{-6}=0.000007$ .

So, the probability of survive, $p=1-q$

$\Rightarrow p=1-7\times 10^{-6}=0.999993.$

(a) The total number of jump he made, n=48

Using Bernoulli's equation, the probability of surviving in exactly 48 jumps (r=48) out of 48 jumps (n=48) is

$=\binom(n,r)p^rq^{n-r}$

$=\binom(48,48)(0.999993)^{48}(0.000007)^{48-48}$

$=(0.999993)^{48}=0.999664$ (approx)

So, the probability of survive in 48 skydiving is 0.999664,

(b) The given probability of surviving =90%=0.9

Let, total $n$ skydiving jumps required to meet the surviving probability of 0.9.

So, By using Bernoulli's equation,

$0.9=\binom {n }{r} p^rq^{n-r}$

Here, r=n.

$\Rightarrow 0.9=\binom{n}{n}p^nq^{n-n}$

$\Rightarrow 0.9=p^n$

$\Rightarrow 0.9=(0.999993)^n$

$\Rightarrow \ln(0.9)=n\ln(0.999993)$ [ taking $\log_e$ both sides]

$\Rightarrow n=\frac {\ln(0.9)}{\ln(0.999993)}$

$\Rightarrow n=15051.45$

The number of diving cant be a fractional value, so bound it to the upper integral value.

Hence, the total number of skydiving required to meet the 90% probability of surviving is 15052.

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