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shutvik [7]
3 years ago
10

Consider the experiment where a pair of fair dice is thrown. Let X denote the random variable whose value is determined by takin

g the maximum of the spots showing on either of the two dice thrown. For example, if a 3 and a 5 were thrown, then X would take the value of Maximum(3,5) = 5. The range of values that X can assume are the positive integers {1,2,3,4,5,6}.
1. Please give the corresponding probabilities for the values of X given below.
Pr(X = 1) =
Pr(X = 2) =
Pr(X = 3) =
Pr(X = 4) =
Pr(X = 5) =
Pr(X = 6) =
2. Further, find the probability that X is divisible by 4. Probability that X is divisible by 4 = ______

Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

1-

P(X=1) = 1/36

P(X=2) = 3/36 = 1/12

P(X=3) = 5/36

P(X=4) = 7/36

P(X=5) = 9/36 = ¼

P(X=6) = 11/36

2-

7/36

Step-by-step explanation:

1-

Attached, you will find a table with the possible values of X.

(See table attached)

As we can see, there are 36 entries so,

<em>P(X=1) = 1/36 </em>

<em> </em>

<em>P(X=2) = 3/36 = 1/12 </em>

<em> </em>

<em>P(X=3) = 5/36 </em>

<em> </em>

<em>P(X=4) = 7/36 </em>

<em> </em>

<em>P(X=5) = 9/36 = ¼ </em>

<em> </em>

<em>P(X=6) = 11/36 </em>

2-

It can be noticed that the only possible value of X which is divisible by 4 is 4 and there are 7 entries in the table, so

P(X divisible by 4) = P(X=4) = 7/36

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Answer:

Null hypothesis:\mu \geq 250        

Alternative hypothesis:\mu < 250  

p_v =P(t_{15}    

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.      

Step-by-step explanation:

1) Data given and notation        

\bar X=510 represent the mean for the sample    

s=50 represent the standard deviation for the sample  

n=16 sample size        

\mu_o =530 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.      

t would represent the statistic (variable of interest)        

p_v represent the p value for the test (variable of interest)    

2) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the claim thet the handle on average is 530 bags per hour:      

Null hypothesis:\mu \geq 530        

Alternative hypothesis:\mu < 530        

We don't know the population deviation and the sample size is lees than 30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:        

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

3) Calculate the statistic        

We can replace in formula (1) the info given like this:        

t=\frac{510-530}{\frac{50}{\sqrt{16}}}=-1.60        

4) Calculate the critical value

We need to begin calculating the degrees of freedom

df=n-1=16-1=15

The critical value for this case would be :

P(t_{15}

The value of a that satisfy this on the normal standard distribution is a=-1.75 and would be the critical value on this case zc=-1.75

5) Calculate the P-value        

Since is a one-side upper test the p value would be:        

p_v =P(t_{15}    

6) Conclusion        

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significantly lower than 530 at 5% of significance.      

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Answer:

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