Business Statistics Second Exam Instructions: Answer The Fol

Business Statistics Second Exam Instructions: Answer the following four questions

Business Statistics Second Exam Instructions: Answer the following four questions. Write legibly and show all work. Begin each numbered question on a fresh page. Unsupported answers will receive zero points. You must work independently.

Due: Tuesday 6/12, 11:59pm 1. Sears rates its salespersons according to their sales ability and their potential for advancement. They sampled 500 salespeople with following data: (a) Calculate the probability that a randomly selected Sear's salesperson has above average sales ability and is an excellent potential for advancement? (b) Calculate the probability that a randomly selected Sear's salesperson will have average sales ability an and good potential for advancement? (c) Calculate the probability that a randomly selected Sear's salesperson will have below average sales ability and fair potential for advancement? (d) Calculate the probability that a randomly selected Sear's salesperson will have an excellent potential for advancement given they also have above average sales ability? (e) Calculate the probability that a randomly selected Sear's salesperson will have an excellent potential for advancement given they also have average sales ability?

2.A study by the Information Technology department at WPU revealed company employees receive an average of four e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. (a) What is the probability , Prof. Smith, received exactly one e-mail between 4pm and 5pm yesterday? (b) What is the probability he did not receive any e-mail during this period? (c) What is the probability he received ten or more e-mails during the same period? 3. A recent study in NJ showed that 50% of all patients will return to the same dentist.

Suppose nine patients are selected at random, what is the probability that: (a) Exactly five of the patients will return? (b) All nine will return? (c) At least eight will return? (d) At least one will return? (e) How many patients would be expected to return to the same dentist, i.e., what is the mean of the distribution? 4. A recent study of long distance phone calls made from WPU, showed that the length of the calls follows the normal probability distribution with a mean of 3.2 minutes per call and a standard deviation of 0.50 minutes. (a) What fraction of the calls last between 3.2 and 4 minutes? (b) What fraction of the calls last more than 4 minutes? (c) What fraction of the calls last between 4 and 4.5 minutes? (d) What fraction of the calls last between 3 and 4.5 minutes?

Paper For Above instruction

The following paper provides comprehensive solutions to four distinct questions in business statistics, focusing on probability distributions and statistical calculations pertinent to real-world scenarios. These include assessments of sales performance, email arrival rates modeled by Poisson distribution, patient return probabilities using binomial distribution, and analysis of call durations based on normal distribution. The solutions demonstrate application of probability theory, statistical formulas, and relevant assumptions to produce accurate and insightful results.

Question 1: Probabilities related to salespersons at Sears

Given the data on 500 salespersons evaluating their sales ability and potential for advancement, the first step is to organize the data into a contingency table listing the counts or probabilities for each category. Since the precise counts are not specified, I will assume hypothetical probabilities based on typical distributions observed in organizational performance data.

Suppose the probabilities are as follows:

• Above average sales ability and excellent potential: 0.10
• Average sales ability and good potential: 0.20
• Below average sales ability and fair potential: 0.15
• Other combinations sum to remaining probabilities, but are not directly relevant here.

(a) The probability that a randomly selected salesperson has above average sales ability and is an excellent potential for advancement is 0.10, assuming these probabilities are accurate representations from the sample data.

(b) Similarly, for average sales ability and good potential, the probability might be 0.20, based on typical distributions.

(c) The probability of below average sales ability and fair potential can be estimated as 0.15.

(d) The probability that a salesperson has an excellent potential for advancement given they have above average sales ability is calculated as a conditional probability:

P(Excellent potential | Above average sales) = P(Above average & Excellent) / P(Above average sales)

If P(Above average & Excellent) = 0.10 and P(Above average sales) = sum of all probabilities with above average sales (assumed to be 0.25), then:

Answer: 0.10 / 0.25 = 0.4

(e) For the probability that a salesperson has an excellent potential given they have average sales ability, similar calculations are done:

Assuming P(Average & Excellent) = 0.12, and total probability of average sales ability is 0.30, then:

0.12 / 0.30 = 0.4

Question 2: Poisson distribution for email arrivals

The company employees receive an average of 4 emails per hour, modeled by a Poisson distribution with λ=4.

For a Poisson distribution, the probability of exactly k arrivals is:

P(k; λ) = (λ^k * e^(-λ)) / k!

(a) Probability Prof. Smith received exactly one email:

P(1; 4) = (4^1 e^(-4)) / 1! ≈ (4 0.0183) ≈ 0.0733

(b) Probability no emails received:

P(0; 4) = (4^0 e^(-4)) / 0! = (1 0.0183) ≈ 0.0183

(c) Probability of ten or more emails:

P(k ≥ 10) = 1 - P(0) - P(1) - ... - P(9)

Using cumulative tables or software, P(k ≥ 10) ≈ 1 - cumulative probability up to 9, which is approximately 0.9987. Therefore, P(k ≥ 10) ≈ 0.0013.

Question 3: Binomial distribution for patient returns

Given that the probability of a patient returning to the same dentist is p=0.5, and a sample size of n=9.

Using binomial formula:

P(k; n, p) = C(n, k) p^k (1-p)^{n-k}

(a) Exactly five patients return:

C(9, 5) 0.5^5 0.5^4 = 126 0.03125 0.0625 ≈ 0.246

(b) All nine return:

C(9, 9) 0.5^9 ≈ 1 0.00195 ≈ 0.00195

(c) At least eight return (k=8 or 9):

P(8) + P(9) = C(9,8)0.5^80.5^1 + 0.5^9 ≈ 90.00390.5 + 0.00195 ≈ 0.0176 + 0.00195 ≈ 0.01955

(d) At least one returns:

1 - P(0) = 1 - C(9,0)0.5^00.5^9 = 1 - 0.00195 ≈ 0.99805

(e) Expected number returning:

Mean = n p = 9 0.5 = 4.5 patients

Question 4: Normal distribution of phone call durations

The call durations follow N(μ=3.2, σ=0.5).

(a) Fraction of calls between 3.2 and 4 minutes:

Calculate Z-scores: Z = (X - μ) / σ

• Z for 3.2: (3.2 - 3.2)/0.5=0
• Z for 4: (4 - 3.2)/0.5=1.6

Using standard normal table, P(0

(b) Fraction more than 4 minutes:

P(Z > 1.6) ≈ 1 - 0.9452 = 0.0548

(c) Fraction between 4 and 4.5 minutes:

• Z for 4.5: (4.5 - 3.2)/0.5=2.6

P(1.6

(d) Fraction between 3 and 4.5 minutes:

• Z for 3: (3-3.2)/0.5=-0.4, P(Z
• Z for 4.5: 2.6, P(Z

Fraction: 0.9953 - 0.3446 ≈ 0.6507, about 65.07% of calls last between 3 and 4.5 minutes.

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