Respuesta :
Answer:
16% of batteries have lifetimes longer than 561 days.
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 = 500, \sigma = 61[/tex]
To the nearest percentage, what percentage of batteries have lifetimes longer than 561 days?
This is 1 subtracted by the pvalue of Z when X = 561. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{561 - 500}{61}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413.
1-0.8413 = 0.1587
Rounding to the nearest percentage, 16% of batteries have lifetimes longer than 561 days.
To the nearest percentage, the percentage of batteries have lifetimes longer than 561 days is 0.1587 approx = 15.87% = 16% approx.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For this case, we can take:
X = the lifetime of the battery in the considered large sample.
Then by the provided information, we get:
[tex]X \sim N(\mu = 500, \sigma = 61)\\[/tex]
The needed probability is P(X > 561)
This can be rewritten as:
[tex]P(X > 561) = 1 - P(X \leq 561)\\[/tex]
Converting this variate to standard normal variate using:
[tex]Z = \dfrac{X -\mu }{\sigma}[/tex], we get this probability as:
[tex]P(X > 561) = 1 - P(X \leq 561) = 1- P(Z \leq z = \dfrac{561 - 500}{61}) \\\\P(X > 561) = 1 - P(Z \leq 1)[/tex]
At Z = 1, the p-value obtained from the z-table is
Thus, we get:
[tex]P(X > 561) = 1 - P(Z \leq 1) = 1- 0.8413 = 0.1587\\[/tex]
Thus, to the nearest percentage, the percentage of batteries have lifetimes longer than 561 days is 0.1587 approx = 15.87% = 16% approx.
Learn more about standard normal distribution here:
 https://brainly.com/question/10984889
