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Anni [7]
2 years ago
9

Many grocery store chains offer customers a card they can scan when they check out and offer discounts to people who do so. To g

et the​ card, customers must give​ information, including a mailing address and​ e-mail address. The actual purpose is not to reward loyal customers but to gather data.
Required:
What data do these cards allow stores to​ gather, and why would they want that​ data?
Mathematics
1 answer:
lana [24]2 years ago
6 0

Answer:

1. The data that these cards allow stores to gather are customer data.

2. The stores want the data to enable them understand the customer, their buying goals, customer preferences, and spending patterns.  These data enable the stores to stock their stores to meet customers needs.  They also use the data to understand the individual customers credit history and buying habits.

Step-by-step explanation:

Customer data are essential assets which businesses use to understand their customers.  They obtain the data through firsthand responses from customers, through investigation, or by asking direct questions.  Many organizations use loyalty cards to help them gather customer data, such as individual preferences and spending patterns.

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A woman is randomly selected from the 18–24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally
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Answer:

X \sim N(114.8,13.1)  

Where \mu=114.8 and \sigma=13.1

We are interested on this probability

P(X>140)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

And we can find this probability using the complement rule:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(114.8,13.1)  

Where \mu=114.8 and \sigma=13.1

We are interested on this probability

P(X>140)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140- 1114.8}{2.6})=P(z>1.924)And we can find this probability using the complement rule:

P(z>1.924)=1-P(z

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Answer:

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Given

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k + 1 = 6 ( subtract 1 from both sides )

k = 5

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3 years ago
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