Question Find the probability that a randomly selected passenger's waiting time is less than $0.75$ minutes, given the waiting times are uniformly distributed between $0$ and $7$ minutes. (Simplify and round to three decimal places.)

Studdy Solution

STEP 1

Assumptions
1. The waiting times are uniformly distributed.
2. The minimum waiting time is 0 minutes.
3. The maximum waiting time is 7 minutes.
4. We are looking for the probability of a waiting time less than 0.75 minutes.

STEP 2

For a uniform distribution, the probability density function (pdf) is constant between the minimum value $a$ and the maximum value $b$ . The pdf is given by:
$f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b$

STEP 3

In this case, $a = 0$ minutes and $b = 7$ minutes. Therefore, the pdf is:
$f(x) = \frac{1}{7 - 0} = \frac{1}{7}$

STEP 4

The probability that a randomly selected passenger has a waiting time less than $x$ minutes is the integral of the pdf from $a$ to $x$ :
$P(X < x) = \int_{a}^{x} f(t) \, dt$

STEP 5

For $x = 0.75$ minutes, we calculate the probability:
$P(X < 0.75) = \int_{0}^{0.75} \frac{1}{7} \, dt$

STEP 6

Integrate the function with respect to $t$ :
$P(X < 0.75) = \left[ \frac{1}{7}t \right]_{0}^{0.75}$

STEP 7

Evaluate the integral at the upper and lower limits:
$P(X < 0.75) = \left[ \frac{1}{7} \times 0.75 \right] - \left[ \frac{1}{7} \times 0 \right]$

STEP 8

Simplify the expression:
$P(X < 0.75) = \frac{0.75}{7}$

STEP 9

Calculate the probability:
$P(X < 0.75) = \frac{0.75}{7} \approx 0.1071$

STEP 10

Round the probability to three decimal places as required:
$P(X < 0.75) \approx 0.107$
The probability that a randomly selected passenger has a waiting time less than 0.75 minutes is approximately 0.107.

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Question Find the probability that a randomly selected passenger's waiting time is less than $0.75$ minutes, given the waiting times are uniformly distributed between $0$ and $7$ minutes. (Simplify and round to three decimal places.)

Studdy Solution

STEP 1

Assumptions
1. The waiting times are uniformly distributed.
2. The minimum waiting time is 0 minutes.
3. The maximum waiting time is 7 minutes.
4. We are looking for the probability of a waiting time less than 0.75 minutes.

STEP 2

For a uniform distribution, the probability density function (pdf) is constant between the minimum value $a$ and the maximum value $b$ . The pdf is given by:
$f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b$

STEP 3

In this case, $a = 0$ minutes and $b = 7$ minutes. Therefore, the pdf is:
$f(x) = \frac{1}{7 - 0} = \frac{1}{7}$

STEP 4

The probability that a randomly selected passenger has a waiting time less than $x$ minutes is the integral of the pdf from $a$ to $x$ :
$P(X < x) = \int_{a}^{x} f(t) \, dt$

STEP 5

For $x = 0.75$ minutes, we calculate the probability:
$P(X < 0.75) = \int_{0}^{0.75} \frac{1}{7} \, dt$

STEP 6

Integrate the function with respect to $t$ :
$P(X < 0.75) = \left[ \frac{1}{7}t \right]_{0}^{0.75}$

STEP 7

Evaluate the integral at the upper and lower limits:
$P(X < 0.75) = \left[ \frac{1}{7} \times 0.75 \right] - \left[ \frac{1}{7} \times 0 \right]$

STEP 8

Simplify the expression:
$P(X < 0.75) = \frac{0.75}{7}$

STEP 9

Calculate the probability:
$P(X < 0.75) = \frac{0.75}{7} \approx 0.1071$

STEP 10

Round the probability to three decimal places as required:
$P(X < 0.75) \approx 0.107$
The probability that a randomly selected passenger has a waiting time less than 0.75 minutes is approximately 0.107.

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Question Find the probability that a randomly selected passenger's waiting time is less than 0.750.750.75 minutes, given the waiting times are uniformly distributed between 000 and 777 minutes. (Simplify and round to three decimal places.)

Studdy Solution

STEP 1

Assumptions1. The waiting times are uniformly distributed.2. The minimum waiting time is 0 minutes.3. The maximum waiting time is 7 minutes.4. We are looking for the probability of a waiting time less than 0.75 minutes.

STEP 2

For a uniform distribution, the probability density function (pdf) is constant between the minimum value a a a and the maximum value b b b. The pdf is given by:f(x)=1b−afora≤x≤b f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b f(x)=b−a1​fora≤x≤b

STEP 3

In this case, a=0 a = 0 a=0 minutes and b=7 b = 7 b=7 minutes. Therefore, the pdf is:f(x)=17−0=17 f(x) = \frac{1}{7 - 0} = \frac{1}{7} f(x)=7−01​=71​

STEP 4

The probability that a randomly selected passenger has a waiting time less than x x x minutes is the integral of the pdf from a a a to x x x:P(X<x)=∫axf(t) dt P(X < x) = \int_{a}^{x} f(t) \, dt P(X<x)=∫ax​f(t)dt

STEP 5

For x=0.75 x = 0.75 x=0.75 minutes, we calculate the probability:P(X<0.75)=∫00.7517 dt P(X < 0.75) = \int_{0}^{0.75} \frac{1}{7} \, dt P(X<0.75)=∫00.75​71​dt

STEP 6

Integrate the function with respect to t t t:P(X<0.75)=[17t]00.75 P(X < 0.75) = \left[ \frac{1}{7}t \right]_{0}^{0.75} P(X<0.75)=[71​t]00.75​

STEP 7

Evaluate the integral at the upper and lower limits:P(X<0.75)=[17×0.75]−[17×0] P(X < 0.75) = \left[ \frac{1}{7} \times 0.75 \right] - \left[ \frac{1}{7} \times 0 \right] P(X<0.75)=[71​×0.75]−[71​×0]

STEP 8

Simplify the expression:P(X<0.75)=0.757 P(X < 0.75) = \frac{0.75}{7} P(X<0.75)=70.75​

STEP 9

Calculate the probability:P(X<0.75)=0.757≈0.1071 P(X < 0.75) = \frac{0.75}{7} \approx 0.1071 P(X<0.75)=70.75​≈0.1071

STEP 10

Round the probability to three decimal places as required:P(X<0.75)≈0.107 P(X < 0.75) \approx 0.107 P(X<0.75)≈0.107The probability that a randomly selected passenger has a waiting time less than 0.75 minutes is approximately 0.107.

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Question Find the probability that a randomly selected passenger's waiting time is less than $0.75$ minutes, given the waiting times are uniformly distributed between $0$ and $7$ minutes. (Simplify and round to three decimal places.)

Assumptions
1. The waiting times are uniformly distributed.
2. The minimum waiting time is 0 minutes.
3. The maximum waiting time is 7 minutes.
4. We are looking for the probability of a waiting time less than 0.75 minutes.

For a uniform distribution, the probability density function (pdf) is constant between the minimum value $a$ and the maximum value $b$ . The pdf is given by:
$f(x) = \frac{1}{b - a} \quad \text{for} \quad a \leq x \leq b$

In this case, $a = 0$ minutes and $b = 7$ minutes. Therefore, the pdf is:
$f(x) = \frac{1}{7 - 0} = \frac{1}{7}$

The probability that a randomly selected passenger has a waiting time less than $x$ minutes is the integral of the pdf from $a$ to $x$ :
$P(X < x) = \int_{a}^{x} f(t) \, dt$

For $x = 0.75$ minutes, we calculate the probability:
$P(X < 0.75) = \int_{0}^{0.75} \frac{1}{7} \, dt$

Integrate the function with respect to $t$ :
$P(X < 0.75) = \left[ \frac{1}{7}t \right]_{0}^{0.75}$

Evaluate the integral at the upper and lower limits:
$P(X < 0.75) = \left[ \frac{1}{7} \times 0.75 \right] - \left[ \frac{1}{7} \times 0 \right]$

Simplify the expression:
$P(X < 0.75) = \frac{0.75}{7}$

Calculate the probability:
$P(X < 0.75) = \frac{0.75}{7} \approx 0.1071$

Round the probability to three decimal places as required:
$P(X < 0.75) \approx 0.107$
The probability that a randomly selected passenger has a waiting time less than 0.75 minutes is approximately 0.107.