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Oduvanchick [21]

3 years ago

5

In a sales effectiveness seminar, a group of sales representatives tried a few approaches to selling a new automobile to a custo

mer: the aggressive approach, the passive approach and a mixture of both. For 1740 customers, the following record was kept:
Sale No Sale Total
Aggressive 265 315 580
Passive 481 99 580
Column 746 414 1160

Suppose a customer is selected at random from the 1160 participating customers. Let us use the following notation for events: A = aggressive approach, Pa-passive approach, S-sale, N = no sale. So, P(A) is the probability that an aggressive approach was used, and so on (a) Compute P(S), P(S I A), and P(S | Pa). (Enter your answers as fractions.)

1 answer:

Ksju [112]3 years ago

8 0

Answer:

1. P(S) = 373/580

2. P(S|A) = 53/116

3. P(S|Pa) = 481/580

Step-by-step explanation:

Given

---------------------Sale ----- No Sale-----Total

Aggressive ----265 --------315 ----------580

Passive ----------481 ---------99 -----------580

Total --------------746---------414------------1160

A = aggressive approach,

Pa = passive approach,

S = sale,

N= no sale.

(a) Computing P(S)

This is calculated as the division of customers that participated in sales by total customers

Customers that participated in sales = 265 + 481 = 746

Total Customers = 1160

P(S) = 746/1160

P(S) = 373/580

b.

P(S|A) means that the probability that a sales occur given that the aggressive method was used.

To solve this, we check the cell where Sales and Aggressive intersect

The cell element = 265

Total = 580

P(S|A) = 265/580

P(S|A) = 265/580

P(S|A) = 53/116

c.

P(S|Pa) means that the probability that a sales occur given that the passive method was used.

To solve this, we check the cell where Sales and Passive intersect

The cell element = 481

Total = 580

P(S|Pa) = 481/580

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Let $$X_1, X_2, ...X_n$$ be uniformly distributed on the interval 0 to a. Recall that the maximum likelihood estimator of a is $

Solnce55 [7]

Answer:

a) $\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

b) $E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

c) $P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

e) On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

Step-by-step explanation:

Part a

For this case we are assuming $X_1, X_2 , ..., X_n \sim U(0,a)$

And we are are ssuming the following estimator:

$\hat a = max(X_i)$

For this case the value for $\hat a$ is always smaller than the value of a, assuming $X_i \sim Unif[0,a]$ So then for this case it cannot be unbiased because an unbiased estimator satisfy this property:

$E(a) - a= 0$ and that's not our case.

Part b

For this case we assume that the estimator is given by:

$E(\hat a) = \frac{na}{n+1}$

And using the definition of bias we have this:

$E(\hat a) - a= \frac{na}{n+1} - a = \frac{na -an -a}{n+1}= \frac{-a}{n+1}$

Since is a negative value we can conclude that underestimate the real value a.

And when we take the limit when n tend to infinity we got that the bias tend to 0.

$\lim_{ n \to\infty} -\frac{1}{n+1}= 0$

Part c

For this case we the followng random variable $Y = max (X_i)$ and we can find the cumulative distribution function like this:

$P(Y \leq y) = P(max(X_i) \leq y) = P(X_1 \leq y, X_2 \leq y, ..., X_n\leq y)$

And assuming independence we have this:

$P(Y \leq y) = P(X_1 \leq y) P(X_2 \leq y) .... P(X_n \leq y) = [P(X_1 \leq y)]^n = (\frac{y}{a})^n$

Since all the random variables have the same distribution.

Now we can find the density function derivating the distribution function like this:

$f_Y (Y) = n (\frac{y}{a})^{n-1} * \frac{1}{a}= \frac{n}{a^n} y^{n-1} , y \in [0,a]$

Now we can find the expected value for the random variable Y and we got this:

$E(Y) = \int_{0}^a \frac{n}{a^n} y^n dy = \frac{n}{a^n} \frac{a^{n+1}}{n+1}= \frac{an}{n+1}$

And the bias is given by:

$E(Y)-a=\frac{an}{n+1} -a=\frac{an-an-a}{n+1}= -\frac{a}{n+1}$

And again since the bias is not 0 we have a biased estimator.

Part e

For this case we have two estimators with the following variances:

$V(\hat a_1) = \frac{a^2}{3n}$

$V(\hat a_2) = \frac{a^2}{n(n+2)}$

On this case we see that the estimator $\hat a_1$ is better than $\hat a_2$ and the reason why is because:

$V(\hat a_1) > V(\hat a_2)$

$\frac{a^2}{3n}> \frac{a^2}{n(n+2)}$

$n(n+2) = n^2 + 2n > n +2n = 3n$ and that's satisfied for n>1.

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Determine if the events A and B are independent. Show all steps.

zmey [24]

<u><em></em></u>

Answer:

$P(A U B ) = \frac{13}{25}$

Step-by-step explanation:

<u><em>STEP:-1</em></u>

Given that events A and B are independent then

P(A ∩B ) = P(A) P(B)

Given that P(A) = 2/5 and P(B) = 1/5

<u><em>STEP:- 2</em></u>

P(A ∩B ) = P(A) P(B)

$P(A U B) = P(A) +P(B) - P(A n B )$

$P(A U B ) = P(A) +P(B) - P(A) P(B)$

<u><em>Step(iii):-</em></u>

<u><em></em></u> $P(A U B ) = \frac{2}{5} +\frac{1}{5} - \frac{2}{5} X \frac{1}{5}$ <u><em></em></u>

<u><em></em></u> $P(A U B ) = \frac{3}{5} - \frac{2}{5} X \frac{1}{5}$ <u><em></em></u>

<u><em></em></u> $P(A U B ) = \frac{3}{5} - \frac{2}{25}$ <u><em></em></u>

<u><em></em></u> $P(A U B ) = \frac{15-2}{25} = \frac{13}{25}$ <u><em></em></u>

<u><em /></u>

<u><em /></u>

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