3 years ago

Hola me ayudan con una tarea soy nuevo

Mathematics

1 answer:

Artemon [7]3 years ago

5 0

Answer:

hola estoy aquí para ayudar

Step-by-step explanation:hola

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3 years ago

Suppose that the value of a stock varies each day from $10.82 to $25.17 with a uniform distribution. Find the third quartile, i.

Zarrin [17]

Answer: Third quartile =$21.58

Step-by-step explanation:

Let x be value of a stock .

For uniform distribution,

probability density function = $\dfrac{1}{b-a}=\dfrac{1}{25.17-10.82}=\dfrac1{14.35}=\dfrac{100}{1435}$

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$\Rightarrow\ \int ^{a}_{10.82}f(x)\ dx=0.75\\\\\Rightarrow\ \int ^{a}_{10.82}\dfrac{100}{1435}\ dx=0.75\\\\\Rightarrow\dfrac{100}{1435}[x]^{a}_{10.82}=0.75\\\\\Rightarrow \dfrac{100}{1435}(a-10.82)=0.75\\\\\Rightarrow a-10.82=0.75\times\dfrac{1435}{100}\\\\\Rightarrow a-10.82=10.7625\\\\\Rightarrow a= 10.7625+10.82= $$21.5825\approx $$21.58$

Hence, 75% of all days the stock is below $21.58 or Third quartile =$21.58 .

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3 years ago

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