Need help with b, picture attached.
1 answer:
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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
(b) the probability that the battery life is less than 10.0 hours
=
(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
Answer:
57.93% probability that a trip will take at least 35 minutes.
Step-by-step explanation:
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:
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:
What is the probability that a trip will take at least 35 minutes
This probability is 1 subtracted by the pvalue of Z when X = 35. So
has a pvalue of 0.4207
1 - 0.4207 = 0.5793
57.93% probability that a trip will take at least 35 minutes.