Respuesta :
Answer:
a) 0.691 = 69.1% probability that a battery lasts more than four hours
b) 25% value = 231
75% value = 299
c) 183 minutes
Step-by-step explanation:
Problems of normally distributed samples are 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]
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:
[tex]\mu = 265, \sigma = 50[/tex]
a) What is the probability that a battery lasts more than four hours?
4 hours = 4*60 = 240 minutes
This is 1 subtracted by the pvalue of Z when X = 240. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{240 - 265}{50}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.309
1 - 0.309 = 0.691
0.691 = 69.1% probability that a battery lasts more than four hours
b) What are the quartiles (the 25% and 75% values) of battery life?
25th percentile:
X when Z has a pvalue of 0.25. So X when Z = -0.675
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = -0.675*50[/tex]
[tex]X = 231[/tex]
75th percentile:
X when Z has a pvalue of 0.75. So X when Z = 0.675
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = 0.675*50[/tex]
[tex]X = 299[/tex]
25% value = 231
75% value = 299
c) What value of life in minutes is exceeded with 95% probability?
The 100-95 = 5th percentile, which is the value of X when Z has a pvalue of 0.05. So X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 265}{50}[/tex]
[tex]X - 265 = -1.645*50[/tex]
[tex]X = 183[/tex]
