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 and standard deviation , the zscore of a measure X is given by
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:
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
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, . So:
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, . So:
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
A value of 131.5 minutes is excedeed with 95% probability.