In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and tH passive approach. For 1160 customers, the following record was kept: Sale No Sale Row Total Aggressive Passive Column Total 252 328 580 475 105 580 727 433 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 | A), and P(S | Pa). (Enter your answers as fractions.) P(S) = P(S | A) = P(S | Pa) = (b) Are the events S = sale and Pa = passive approach independent? Explain. O No. The two events cannot occur together. O Yes. P(S) = P(S | Pa). ONo. P(S) # P(S | Pa). O Yes. The two events can occur together. (c) Compute P(A and S) and P(Pa and S). (Enter your answers as fractions.) P(A and S) = P(Pa and S) =

MATLAB: An Introduction with Applications
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**Educational Content on Sales Effectiveness Using Statistical Analysis**

In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept:

|               | Sale | No Sale | Row Total |
|---------------|------|---------|-----------|
| **Aggressive**| 252  | 328     | 580       |
| **Passive**   | 475  | 105     | 580       |
| **Column Total**   | 727  | 433     | 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 \( P(S) \), \( P(S \mid A) \), and \( P(S \mid Pa) \). (Enter your answers as fractions.)

\( P(S) = \underline{\phantom{ANSWER}} \)

\( P(S \mid A) = \underline{\phantom{ANSWER}} \)

\( P(S \mid Pa) = \underline{\phantom{ANSWER}} \)

(b) Are the events \( S = \) sale and \( Pa = \) passive approach independent? Explain.

- ☐ No. The two events cannot occur together.
- ☐ Yes. \( P(S) = P(S \mid Pa) \).
- ☐ No. \( P(S) \neq P(S \mid Pa) \).
- ☐ Yes. The two events can occur together.

(c) Compute \( P(A \text{ and } S) \) and \( P(Pa \text{ and } S) \). (Enter your answers as fractions.)

\( P(A \text{ and } S) = \underline{\phantom{ANSWER}} \)

\( P(Pa \text{ and } S) = \underline{\phantom{ANSWER}} \)

(d) Compute \( P(N) \) and \( P(N \mid A) \). (Enter your answers as fractions.)

\( P(N) = \underline{\phantom{ANSWER}} \)

\( P(N
Transcribed Image Text:**Educational Content on Sales Effectiveness Using Statistical Analysis** In a sales effectiveness seminar, a group of sales representatives tried two approaches to selling a customer a new automobile: the aggressive approach and the passive approach. For 1160 customers, the following record was kept: | | Sale | No Sale | Row Total | |---------------|------|---------|-----------| | **Aggressive**| 252 | 328 | 580 | | **Passive** | 475 | 105 | 580 | | **Column Total** | 727 | 433 | 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 \( P(S) \), \( P(S \mid A) \), and \( P(S \mid Pa) \). (Enter your answers as fractions.) \( P(S) = \underline{\phantom{ANSWER}} \) \( P(S \mid A) = \underline{\phantom{ANSWER}} \) \( P(S \mid Pa) = \underline{\phantom{ANSWER}} \) (b) Are the events \( S = \) sale and \( Pa = \) passive approach independent? Explain. - ☐ No. The two events cannot occur together. - ☐ Yes. \( P(S) = P(S \mid Pa) \). - ☐ No. \( P(S) \neq P(S \mid Pa) \). - ☐ Yes. The two events can occur together. (c) Compute \( P(A \text{ and } S) \) and \( P(Pa \text{ and } S) \). (Enter your answers as fractions.) \( P(A \text{ and } S) = \underline{\phantom{ANSWER}} \) \( P(Pa \text{ and } S) = \underline{\phantom{ANSWER}} \) (d) Compute \( P(N) \) and \( P(N \mid A) \). (Enter your answers as fractions.) \( P(N) = \underline{\phantom{ANSWER}} \) \( P(N
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