Mathematics, 07.04.2020 05:01, Ashu638

A company that manufactures smartphones developed a new battery that has a longer life span than that of a traditional battery. From the date of purchase of a smartphone, the distribution of the life span of the new battery is approximately normal with mean 30 months and standard deviation 8 months. For the price of $50, the company offers a two-year warranty on the new battery for customers who purchase a smartphone. The warranty guarantees that the smartphone will be replaced at no cost to the customer if the battery no longer works within 24 months from the date of purchase.
1. in how many months from the date of purchase is it expected that 25 percent of the batteries will no longer work? Justify your answer.
2. Suppose one customer who purchases the warranty is selected at random. What is the probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works?

Answers: 2

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Answers

answered: mariobros3090

1) In 24.6 months from the date of purchase it is expected that 25 percent of the batteries will no longer work.

2) There is 22.66% probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works.

Step-by-step explanation:

Given :

Mean -- $\rm \mu = 30 \; months$

Standard deviation -- $\rm \sigma = 8\; months$

Calculation :

This is the normally distributed sample problem and is solved by using the z-score formula.

$z_s_c_o_r_e = \dfrac{x-\mu}{\sigma}$

1) This is the 25th percentile of duration, which is the value of x when z has a p-value of 0.25. So it is x when z = -0.675.

$-0.675=\dfrac{x-30}{8}$

$\rm x = 24.6 \; months$

Therefore, in 24.6 months from the date of purchase it is expected that 25 percent of the batteries will no longer work.

2) When x = 24,

$z_s_c_o_r_e = \dfrac {24 - 30}{8}$

z = - 0.75

has a pvalue of 0.2266

Therefore, there is 22.66% probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works.

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answered: lissa34

1. 24.6 months, which is the value of X when Z has a pvalue of 0.25

2. 22.66% probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 30, \sigma = 8$

1. in how many months from the date of purchase is it expected that 25 percent of the batteries will no longer work? Justify your answer.

This is the 25th percentile of duration, which is the value of X when Z has a pvalue of 0.25. So it is X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 30}{8}$

X - 30 = -0.675*8

X = 24.6

24.6 months, which is the value of X when Z has a pvalue of 0.25

2. Suppose one customer who purchases the warranty is selected at random. What is the probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works?

This is the pvalue of Z when X = 24. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24 - 30}{8}$

Z = -0.75

Z = -0.75 has a pvalue of 0.2266

22.66% probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works

answered: Guest

step-by-step explanation:

first, eliminate the parenthesis:

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second, group like terms:

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third, combine like terms:

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