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vivado [14]
3 years ago
10

The probability that a student at your school takes Drivers Education and Spanish is 87/1000. The probability that a student tak

es Spanish is 68/100. What is the probability that a student takes Drivers Ed given that the Student is taking Spanish?
Mathematics
2 answers:
Gnom [1K]3 years ago
5 0
Let
x---------> <span>The probability that a student takes Spanish 
y-------> </span><span>the probability that a student takes Drivers Education given that the Student is taking Spanish
z-------> </span><span>The probability that a student at school takes Drivers Education and Spanish 

we know that
z=x*y------> solve for y
y=z/x
z=87/1000
x=68/100
substitute
y=(87/1000)/(68/100)-----------> y=87/680

the answer is
87/680</span>
TEA [102]3 years ago
5 0

Answer: The probability is P = 0.128

Step-by-step explanation:

The data we have is:

The probability of a student to take drivers education and Spanish is 87/1000.

The probability that a student takes Spanish is 68/100.

Now, remember that if for event 1 we have the probability p1, and for event 2 we have the probability p2, the probability of both events happening is:

P = p1*p2

This is the case for the student that takes the two classes, but when we assume that the student takes Spanish, we can remove the probability of that event (because we are already looking at the 68/100 of the cases where the student selected Spanish)

So given that a student is tanking Spanish, the probability of him to take drivers ed is:

(Probability of both classes)/(probability of tanking Spanish)

P = (87/1000)*(100/68) = 0.128

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An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 180180 engin
tiny-mole [99]

Answer:

z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98    

p_v =P(Z  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.  

Step-by-step explanation:

Data given and notation  

\bar X=4.6 represent the sample mean

\sigma=0.9 represent the population standard deviation

n=180 sample size  

\mu_o =4.8 represent the value that we want to test

\alpha=0.1 represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)  

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

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if thetrue mean is below the specifications, the system of hypothesis would be:  

Null hypothesis:\mu \geq 4.8  

Alternative hypothesis:\mu < 4.8  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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

z-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".  

Calculate the statistic

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

z=\frac{4.6-4.8}{\frac{0.9}{\sqrt{180}}}=-2.98    

P-value

Since is a one sided test the p value would be:  

p_v =P(Z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis,and we have enough evidence to conclude that the true mean is significantly lower thn 4.8 at 10% of signficance.  

4 0
3 years ago
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