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2 years ago

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A company administers an "aptitude test for managers" to aid in selecting new management trainees. Prior experience suggests tha

t 60 percent of all applications for management trainee positions would be successful if they were hired. Furthermore, past experience with the aptitude test indicates that 85 percent of applicants who turn out to be successful managers pass the test and 90 percent of applicants who do not turn out to be successful managers fail the test.

1 answer:

RoseWind [281]2 years ago

5 0

Answer:

0.94

Step-by-step explanation:

The question after this basically is:

<em>"If the applicant passes the "aptitude test for managers", what is the probability that the applicant will succeed in the management position?"</em>

<em />

So,

P(successful if hired) = 60% = 0.6 [let it be P(x)]

P(success at passing the test) = 85% = 0.85 [let it be P(y)]

P(successful and pass the test) = P(x) + P(y) -[P(x)*P(y)]

So,

P(successful and pass the test) = 0.6 + 0.85 - (0.6*0.85) = 0.94 (94%)

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So the value of height that separates the bottom 75% of data from the top 25% is 236.121.

b) $P(X \geq 3) = 1-P(X$

c) $P(\bar X \geq 225)=1- P(\bar X$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the cuts of a population, and for this case we know the distribution for X is given by:

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Where $\mu=225$ and $\sigma=16.5$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.25$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.674=\frac{a-225}{16.5}$

And if we solve for a we got

$a=225 +0.674*16.5=236.121$

So the value of height that separates the bottom 75% of data from the top 25% is 236.121.

Part b

For this case we know that the individual probability of select one wheel with a cutting rate higher than the calculated value in part a is 0.25, and we select n =10 so then we can use the binomial distribution for this case:

$X\sim Bin(n=10, p=0.25)$

And we want this probability:

$P(X \geq 3) = 1-P(X$

We can find the individual probabilities like this:

$P(X=0)=(10C0)(0.25)^0 (1-0.25)^{10-0}=0.0563$

$P(X=1)=(10C1)(0.25)^1 (1-0.25)^{10-1}=0.1877$

$P(X=2)=(10C2)(0.25)^2 (1-0.25)^{10-2}=0.2816$

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Part c

For this case we know that the distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want this probability:

$P(\bar X \geq 225)$

And for this case we can use the complement rule and the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we replace we got:

$P(\bar X \geq 225)=1- P(\bar X$

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