Using the rules for significant figures what do you get when you subtract 12.63 from 52.2147
1 answer:
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Answer:
Probability that a battery will last more than 19 hours is 0.0668.
Step-by-step explanation:
We are given that the lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.
<em>Let X = lifetime of a battery in a certain application</em>
So, X ~ N()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean lifetime = 16 hours
= standard deviation = 2 hours
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.
So, the probability that a battery will last more than 19 hours is given by = P(X > 19 hours)
P(X > 19) = P( >
) = P(Z > 1.50) = 1 - P(Z
1.50)
= 1 - 0.9332 = 0.0668
<em>Now, in the z table the P(Z </em><em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.</em>
Hence, the probability that a battery will last more than 19 hours is 0.0668.