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g100num [7]
2 years ago
13

A random number is selected from the interval [6.35, 10]. Find the probability that the number is within a distance of 0.25 from

an even integer. (Answer as a decimal number, and round to 4 decimal places).
Mathematics
1 answer:
exis [7]2 years ago
6 0

Let <em>X</em> be a random number selected from the interval. Then the probability density for the random variable <em>X</em> is

f_X(x)=\begin{cases}\dfrac1{10-6.35}=\dfrac1{3.65}\approx0.2740&\text{if }6.35\le x\le 10\\0&\text{otherwise}\end{cases}

8 and 10 are the only even integers that fit the given criterion (6 is more than 0.25 away from 6.35), so that we're looking to compute

P(|<em>X</em> - 8| < 0.25) + P(|<em>X</em> - 10| < 0.25)

… = P(7.75 < <em>X</em> < 8.25) + P(9.75 < <em>X</em> < 10.25)

… = P(7.75 < <em>X</em> < 8.25) + P(9.75 < <em>X</em> < 10)

(since P(<em>X</em> > 10) = 0)

… = 0.2740 (8.25 - 7.75) + 0.2740 (10 - 9.75)

… = 0.2055

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A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

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Step-by-step explanation:

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