A phone manufacturer offers its customers two different versions of the same phone. The low storage version costs $350, and the high storage version costs $450. let X represent the manufacturer's income on a randomly chosen purchase of this phone. Based on previous data, here's the probability distribution of X along with summary statistics:

Low capacity High capacity
х income ($) 350 450
P(X) 0.65 0.35
Mean: μx=$385
Standard deviation: σx= $48
The manufacturer is considering a $25 price increase on both versions of the phone. Assume that this increase will not change the probability that corresponds to each version. Let Y represent their income on a randomly chosen purchase of this higher priced phone. What are the mean and standard deviation of Y?

Answers

Answer 1

Answer:

Mean = $410

Standard deviation = $48

Step-by-step explanation:

From the given information:

                              Low capacity         High capacity

Y income ($)             350 + 25                450  + 25

                                       = 375                    = 475

P(Y)                                0.65                      0.35

[tex]U_Y = (375 \times 0.65) + (475 \times 0.35)[/tex]

[tex]U_Y = 243.75 + 166.25[/tex]

[tex]\mathbf{U_Y = 410}[/tex]

[tex]Var _{(Y)} = E(Y^2) - E(Y)^2[/tex]

[tex]Var _{(Y)} = \Big [(375)^2 \times 0.65 + (475)^2 \times 0.35 \Big ]- (410)^2[/tex]

[tex]Var _{(Y)} =170375 - 168100[/tex]

[tex]Var _{(Y)} = 2275[/tex]

[tex]\sigma _Y = \sqrt{Var _{(Y)}}[/tex]

[tex]\sigma _Y = \sqrt{2275}[/tex]

[tex]\mathbf{\sigma _Y \simeq 48}[/tex]

Answer 2

This question is based on the concept of statistics. Therefore, the mean of Y is [tex]U_y[/tex] = 410 and the standard deviation of Y is  [tex]\sigma_{(Y)}[/tex] =48.

Given:

Low capacity = $350

High capacity = $450

P(X)  of low capacity = 0.65

P(X)  of high capacity = 0.35

Mean: μx=$385

Standard deviation: σ(x)= $48

We need to calculate the the mean and standard deviation of Y (Y represent their income on a randomly chosen purchase of this higher priced phone).

According to the question,

From the given information, it is observe that,

                             Low capacity         High capacity

Y income ($)             350 + 25                450  + 25

                                      = 375                    = 475

P(Y)                               0.65                      0.35

Now, calculate the mean of Y,

[tex]U_y = (375 \times 0.65) + ( 475 \times 0.35)\\U_y = 243.75 + 166.25\\\bold{U_y = 410}[/tex]

Now, calculate the variance of Y,

[tex]Var_{(Y)} = E( Y^2) - E ( Y) ^2\\\\Var_{(Y)} = [(375)^2 \times 0.65 + (475)^2 \times 0.35] - (410)^2\\\\Var_{(Y)} = 170375 - 168100\\\\Var_{(Y)} = 2275\\[/tex]

Calculating the standard deviation of Y,

[tex]\sigma_{(Y)} = \sqrt{Var_{(Y)}} \\\\\sigma_{(Y)} = \sqrt{2275} \\\\\sigma_{(Y)} = 47.6 \approx 48[/tex]

Therefore, the mean of Y is [tex]U_y[/tex] = 410 and the standard deviation of Y is  [tex]\sigma_{(Y)}[/tex] =48.

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