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Dafna1 [17]
3 years ago
15

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean va

lue 10 min and standard deviation 2 min. if five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?
Mathematics
1 answer:
xz_007 [3.2K]3 years ago
3 0

To solve for the probability proportion, we make use of the z statistic. The procedure to do is to calculate for the z value and using the standard probability tables, we can look up for the p value. The formula for z score is:

z =(x – μ) / (σ / sqrt(n))

where,

x = sample score = 11

μ = sample mean= 10

σ = standard deviation = 2

n = sample size

 

Calculating for the z and p value when n = 5:

z =(11 – 10) / (2 / sqrt(5))

z = 1.12

Using the tables, p(5) = 0.8686

 

Calculating for the z and p value when n = 6:

z =(11 – 10) / (2 / sqrt(6))

z = 1.22

Using the tables, p(6) = 0.8888

 

If both days should be occuring, therefore the total probability that each day is at most 11 min is:

p total = p(5) * p(6)

p total = 0.8686 * 0.8888

p total = 0.772

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The probability that a student has a Visa card (event V) is .73. The probability that a student has a MasterCard (event M) is .1
snow_lady [41]

We assumed in this answer that the question b is, Are the events V and M independent?

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Step-by-step explanation:

The key factor to solve these questions is to know that:

\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)

We already know from the question the following probabilities:

\\ P(V) = 0.73

\\ P(M) = 0.18

The probability that a student has both cards is 0.03. It means that the events V AND M occur at the same time. So

\\ P(V \cap M) = 0.03

The probability that a student has either a Visa card or a MasterCard

We can interpret this probability as \\ P(V \cup M) or the sum of both events; that is, the probability that one event occurs OR the other.

Thus, having all this information, we can conclude that

\\ P(V \cup M) = P(V) + P(M) - P(V \cap M)

\\ P(V \cup M) = 0.73 + 0.18 - 0.03

\\ P(V \cup M) = 0.88

Then, <em>the probability that a student has either a Visa card </em><em>or</em><em> a MasterCard is </em>\\ P(V \cup M) = 0.88.<em> </em>

Are the events V and M independent?

A way to solve this question is by using the concept of <em>conditional probabilities</em>.

In Probability, two events are <em>independent</em> when we conclude that

\\ P(A|B) = P(A) [1]

The general formula for a <em>conditional probability</em> or the probability that event A given (or assuming) the event B is as follows:

\\ P(A|B) = \frac{P(A \cap B)}{P(B)}

If we use the previous formula to find conditional probabilities of event M given event V or vice-versa, we can conclude that

\\ P(M|V) = \frac{P(M \cap V)}{P(V)}

\\ P(M|V) = \frac{0.03}{0.73}

\\ P(M|V) \approx 0.041

If M were independent from V (according to [1]), we have

\\ P(M|V) = P(M) = 0.18

Which is different from we obtained previously;

That is,

\\ P(M|V) \approx 0.041

So, the events V and M are not independent.

We can conclude the same if we calculate the probability

\\ P(V|M), as follows:

\\ P(V|M) = \frac{P(V \cap M)}{P(M)}

\\ P(V|M) = \frac{0.03}{0.18}

\\ P(V|M) = 0.1666.....\approx 0.17

Which is different from

\\ P(V|M) = P(V) = 0.73

In the case that both events <em>were independent</em>.

Notice that  

\\ P(V|M)*P(M) = P(M|V)*P(V) = P(V \cap M) = P(M \cap V)

\\ \frac{0.03}{0.18}*0.18 = \frac{0.03}{0.73}*0.73 = 0.03 = 0.03

\\ 0.03 = 0.03 = 0.03 = 0.03

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