Answer:
0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
![\mu = 606, \sigma = 62, n = 99, s = \frac{62}{\sqrt{99}} = 6.23](https://tex.z-dn.net/?f=%5Cmu%20%3D%20606%2C%20%5Csigma%20%3D%2062%2C%20n%20%3D%2099%2C%20s%20%3D%20%5Cfrac%7B62%7D%7B%5Csqrt%7B99%7D%7D%20%3D%206.23)
What is the probability that the mean battery life would be greater than 619 minutes?
This is 1 subtracted by the pvalue of Z when X = 619. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{619 - 606}{6.23}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B619%20-%20606%7D%7B6.23%7D)
![Z = 2.09](https://tex.z-dn.net/?f=Z%20%3D%202.09)
has a pvalue of 0.9817
1 - 0.9817 = 0.0183
0.0183 = 1.83% probability that the mean battery life would be greater than 619 minutes