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svp [43]
3 years ago
11

A popular clothing brand has released a new range of pants. The brand manager wants to target several sections of youth with thi

s range. So the
manager divides the target group into four types: trendy, formal, budget, and comfort. Which application of data mining is the manager using?
A.
customer classification
B.
customer retention
c.
fraud detection
D.
direct marketing
Business
1 answer:
Andrej [43]3 years ago
7 0

.Answer:

A.  customer classification

Explanation:

Classification of consumers is the process of grouping customers according to shared traits. Customers in the same group will share some common characteristics that a business can use to serve them better. In customer classification, the firm seeks to identify the common traits that make customers have similar buying patterns.

The manager in the clothing brand has identified traits he can use to classify the target customers into four groups. He has applied customer classification. If he subdivides each group by specific attributes such as age, gender, or other similarities, he would be doing customer segmentation

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The mean ls the average of a distribution obtained by taking the ratio of the sum and the frequency. The mean and standard deviation values of the distribution are 3.30 and 1.26 respectively.

We know,

  • Mean, μ = ΣX/ n
  • Standard deviation = √variance

To calculate mean and variance :-

X = 2,3,2,3,3,4,2,4,3,2,2,7,3,4,5,3,3,3,3,5

  • ΣX = sum of X = 66
  • n = sample size = 20

As, Mean = sum of x / sample size

66 / 20 = 3.3

Mean = 3.3

The variance = [Σ(X - μ)²] ÷ (n - 1)

Variance = [Σ(X - μ)²] ÷ (n - 1)

= (30.2) / 19 = 1.589

The standard deviation = √variance

= √1.589

= 1.26

Standard deviation =  1.26

Therefore,

The mean and standard deviation values are 3.3 and 1.26.

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Correct Question - Scrapper Elevator Company has 20 sales representatives who sell its product throughout the United States and Canada. The number of units sold last month by each representative is listed below. Assume these sales figures to be the population values. 2,3,2,3,3,4,2,4,3,2,2,7,3,4,5,3,3,3,3,51. Compute the mean of the population.2. Compute the standard deviation.3. what would be the mean of the sample means?4. what would be the standard deviation of the sample means?

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Answer:

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Explanation:

As for the provided information we have:

Par value = $1,000

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Effective interest rate semiannually = 11%/2 = 5.5% = 0.055

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