The time until recharge for a battery in a laptop computer under common conditions is normally distributed with a mean of 260 minutes and a standard deviation of 50 minutes. a. What is the probability that a battery lasts more than four hours? b. What are the quartiles (the 25% and 75% values) of battery life? c. What value of life in minutes is exceeded with 95% probability? 4.5.7 .WP . SS V

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Answer:

There is a 65.54% probability that the battery lasts more than four hours.

The first quartile is 226.5 minutes.

The third quartile is 294 minutes.

A value of 131.5 minutes is excedeed with 95% probability.

Step-by-step explanation:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

In this problem, we have that:

[tex]\mu = 260, \sigma = 50[/tex]

a. What is the probability that a battery lasts more than four hours?

4 hours is 4*60 = 240 minutes. The pvalue of the Zscore of X = 240 is the probability that the battery lasts less than four hours. The subtraction of 100% by this value is the probability that the battery lasts more than four hours.

So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{240 - 260}{50}[/tex]

[tex]Z = -0.4[/tex]

A Zscore of -0.4 has a pvalue of 0.3446. This means that there is a 34.46% probability that the battery lasts less than four hours.

The probability that the battery lasts more than four hours is 1 - 0.3446 = 0.6554.

There is a 65.54% probability that the battery lasts more than four hours.

b. What are the quartiles (the 25% and 75% values) of battery life?

The first quartile is the value of X when Z has a pvalue of 0.25.

When the pvalue is 0.25, [tex]Z = -0.67[/tex]. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.67 = \frac{X- 260}{50}[/tex]

[tex]X - 260 = -33.5[/tex]

[tex]X = 226.5[/tex]

The first quartile is 226.5 minutes.

The third quartile is the value of X when Z has a pvalue of 0.75.

When the pvalue is 0.75, [tex]Z = 0.68[/tex]. So:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.68 = \frac{X- 260}{50}[/tex]

[tex]X - 260 = 34[/tex]

[tex]X = 294[/tex]

The third quartile is 294 minutes.

What value of life in minutes is exceeded with 95% probability?

This is the value of X when the Zscore has a pvalue of 0.05. That is [tex]Z = -2.57[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-2.57 = \frac{X- 260}{50}[/tex]

[tex]X - 260 = -128.5[/tex]

[tex]X = 131.5[/tex]

A value of 131.5 minutes is excedeed with 95% probability.

A) To derive the probability that the battery lasts more than fours hours, we need to find a Z score. The Z score and apply the normal probability Z table to get the probability. We use the normal table because the process of charging and discharging follows a normal distribution pattern.

Z, therefore, = (X - Mean)/Standard Deviation, where

X = 4 hours or (240 Minutes)

Thus, Z = [tex]\frac{(240-260)}{50}[/tex]

= - 0.4

We can thus say that:

P(Z> -0.4) = P(Z < 0.4) = 0.66

B) to establish the quantiles,  we need to determine the value of Z for the lower 25% and the upper 25% of the data. The upper is -0.67 and the lower is 0.67.

Mean + Z (Standard Deviation)

This translates to:

250 + (-0.67)(50)

= 215.50 (25%);

250 + (0.67)(50)

= 283.50 (75%)

C) The value of life in minutes that is exceeded with 95% probability is ascertained as follows

250 +  (-1.65)(50) = 167.50

What is Normal Distribution?

This refers to the concept which is also known as the Gaussian distribution. A data sample in statistics is said to be normally distributed if its probabilty distribution is symmetric about the mean point of that data.

See the link below for more exercises related to Normal Distribution:

https://brainly.com/question/4079902:

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