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serg [7]
3 years ago
5

The random variable X is exponentially distributed, where X represents the time it takes for a person to choose a birthday gift.

If X has an average value of 26 minutes, what is the probability that X is less than 30 minutes? (Do not round until the final step. Round your answer to 3 decimal places.)
Mathematics
1 answer:
Karolina [17]3 years ago
8 0

Answer:

0.685 = 68.5% probability that X is less than 30 minutes

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

If X has an average value of 26 minutes

This means that m = 26, \mu = \frac{1}{26}

What is the probability that X is less than 30 minutes?

P(X \leq 30) = 1 - e^{-\frac{30}{26}} = 0.685

0.685 = 68.5% probability that X is less than 30 minutes

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Leviafan [203]

Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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kolezko [41]

C. 6

would be the answer

8 0
3 years ago
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