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Alona [7]
3 years ago
5

Volten Inc. is an apparel store. It has separate sections for casual wear, office wear, and party wear. It offers a broad select

ion of clothing across various price ranges to accommodate consumers at different income levels. In this way, Volten caters to different socioeconomic groups. Volten targets its market segments using _____ as its base for product positioning.
a. product class
b. application
c. competitor
d. product user
Business
1 answer:
yarga [219]3 years ago
4 0

Answer:

The answer is: D) product user

Explanation:

Product user segmentation refers to the marketing practice of dividing potential customers into segments based on the characteristics they might share, for example: shared behavior, workplace, leisure activities, hobbies, etc.

Usually the better you know your customers, the better you can divide them into segments, although one single customer may apply to all of them.

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Credit card debt. The average credit card debt for college seniors in $3262. If the debt is normally distributed with a standard
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Answer:

  • 0.98 approximately
  • 0.25 approximately
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Explanation:

standard deviation by which debt is distributed  = $1100

average credit card debt for college seniors = $3262

A)  probability that senior owes at least $1000

P (x ≥ 1000 ) = P ( Z ≥ \frac{1000- \alpha }{\beta }  )

where \alpha = average credit card debt = $3262

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Z = random variable representing credit card debt for college seniors

back to equation P ( Z ≥ \frac{1000- 3262}{1100} )

therefore  P ( z ≥ -2.06 )

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therefore probability of the senior owing $1000 = 1 - 0.0199 = 0.9801

B) probability that senior owes more than $4000

p ( x ≥ 4000) = P ( Z ≥ \frac{4000 - \alpha }{\beta } )

                     = P ( Z ≥ \frac{4000 - 3262}{1100} )

                     = 1 - p ( Z ≤ 0.67 ) therefore probability that senior owes more than $4000 = 1 - 0.7468 = 0.2514

C ) Probability that the senior owes between $300 and $4000

P ( 3000 ≤ x ≤ 4000) = P ( \frac{3000 - \alpha }{\beta } ≤ Z ≤ \frac{4000 - \alpha }{\beta } )

                                  = P ( \frac{3000 - 3262}{1100} ≤ z ≤ \frac{4000 - 3262}{1100} )

                                  = P ( - 0.24 ≤ z ≤ 0.67 )  

                                 = p ( z ≤ 0.67 ) - p ( z ≤ - 0.24 )        

                                 = 1 - p ( z ≥ 0.67 ) - 1 - p ( z ≥ -0.24 )

                                 = 0.7846 - 0.4052 = 0.3434                  

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