Respuesta :
Answer:
Step-by-step explanation:
given that a laptop company claims up to 11.0 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 10.5 hours and standard deviation 27 minutes.
convert into same units into hours.
X is N(10.5, 0.45)
a) the probability that the battery life is at least 11.0 hours
[tex]P(X\geq 11)\\\\=1-0.8667\\=0.1333[/tex]
(b) the probability that the battery life is less than 10.0 hours
=[tex]P(X<10) \\= 0.1333[/tex]
(c) the time of use that is exceeded with probability 0.97
=97th percentile
= 11.844
d) The time of use that is exceeded with probability 0.9 is
is 90th percentile = 10.885
Z-table is known as the standard normal distribution table. The probability of probability that the battery life is less than 10.0 hours is 13.35%.
What is a Z-table?
A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.
As it is given that the mean of the battery life is 10.5 hours, while the standard deviation of the battery life is 27 minutes, therefore, the mean and the standard deviation of the battery can be written as,
[tex]\rm \text{Standard Deviation},(\sigma) = 27\ minutes\\\\\text{Mean},(\mu) = 10.5 hours = 10.5 \times 60 = 630\ minutes[/tex]
A.) We know the mean and the standard deviation of the life of the battery, therefore, the probability that the battery life is at least 11.0 hours can be written as,
[tex]\begin{aligned}P(X \geq 11\ hours) &= 1 - P(X < 11\ hours)\\P(X \geq 11\ hours) &= 1 - P(z < \dfrac{10-\mu}{\sigma})\\P(X\geq 660\ minutes) &=1- P(z \leq \dfrac{660-630}{27})\\\\\end{aligned}[/tex]
[tex]\begin{aligned}P(X\geq 11\ hours) &= 1-0.8665\\ \\&=13.35\% \end{aligned}[/tex]
Hence, the probability that the battery life is at least 11.0 hours is 13.35%.
B.) We know the mean and the standard deviation of the life of the battery, therefore, the probability of the probability that the battery life is less than 10.0 hours can be written as,
[tex]\begin{aligned}P(X < 10\ hours) &= P(X\leq 10\ hours)\\P(X\leq 10\ hours) &= P(z \leq \dfrac{10-\mu}{\sigma})\\P(X\leq 600\ minutes) &=P(z \leq \dfrac{600-630}{27})\\\\\end{aligned}[/tex]
[tex]\begin{aligned}P(X\leq 10\ hours) &= 0.1335\\ \\&=13.35\% \end{aligned}[/tex]
Hence, the probability that the battery life is less than 10.0 hours is 13.35%.
C.) We know the mean and the standard deviation of the life of the battery, therefore, the time of use that is exceeded with probability 0.97 can be written as,
[tex]P(X\geq x) = 0.97\\\\\text{From the Z-table}\\\\z = \dfrac{x-\mu}{\sigma} = 1.88\\\\x = 650.76\ {\rm minutes}\\\\x = 10.85 {\rm\ hours}[/tex]
Hence, the time of use that is exceeded with a probability of 0.97 is 10.85 hours.
Learn more about Z-table:
https://brainly.com/question/6096474
