Question

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

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

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

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:

$\mu = 260, \sigma = 50$

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

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

$Z = \frac{240 - 260}{50}$

$Z = -0.4$

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, $Z = -0.67$. So:

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

$-0.67 = \frac{X- 260}{50}$

$X - 260 = -33.5$

$X = 226.5$

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, $Z = 0.68$. So:

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

$0.68 = \frac{X- 260}{50}$

$X - 260 = 34$

$X = 294$

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 $Z = -2.57$

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

$-2.57 = \frac{X- 260}{50}$

$X - 260 = -128.5$

$X = 131.5$

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

Answer 2

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 = $\frac{(240-260)}{50}$

= - 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.

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