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Alenkinab [10]
3 years ago
8

Algebra Help please?

Mathematics
1 answer:
tresset_1 [31]3 years ago
4 0

According to the table, f(x)=g(x) when x=0 because f(0)=g(0)=1, and when x=1 because f(1)=g(1)=0.25.

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Round 128.59943 to 3 decimal places
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128.59943

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(Uniform) Suppose X follows a continuous uniform distribution from 4 to 11. (a) Write down the PDF of X. (b) Find P(X ≤ 7). Roun
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Answer:

a) f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

b) P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

c) P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

d) P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

Step-by-step explanation:

For this case we assume that the random variable X follows this distribution:

X \sim Unif (a=4, b =11)

Part a

The probability density function is given by the following expression:

f(x) = \frac{1}{b-a} , a \leq x \leq b

f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11

Part b

We want this probability:

P(X \leq 7)

And we can use the cumulative distribution function given by:

F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4}

And replacing we got:

P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286

Part c

We want this probability:

P(5 < X \leq 7)

And we can use the CDF again and we have:

P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857

Part d

We want this conditional probabilty:

P(X >5 | X \leq 7)

And we can find this probability with this formula from the Bayes theorem:

P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5

7 0
3 years ago
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