Mathematics, 18.01.2021 14:00 pevin2955

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?

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Answer from: ayoismeisjuam

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

$U_Y = (375 \times 0.65) + (475 \times 0.35)$

U_Y = 243.75 + 166.25

$\mathbf{U_Y = 410}$

$Var _{(Y)} = E(Y^2) - E(Y)^2$

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

$Var _{(Y)} =170375 - 168100$

$Var _{(Y)} = 2275$

$\sigma _Y = \sqrt{Var _{(Y)}}$

$\sigma _Y = \sqrt{2275}$

$\mathbf{\sigma _Y \simeq 48}$

Answer from: Quest

step-by-step explanation:

i found the life tables for the year 2012.

life table of males at age 30. probability of dying between ages x and x+1

total population:

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old.

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

Answer from: Quest

The last one bc that's a calendar

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