Respuesta :
Answer with Step-by-step explanation:
The life of a certain brand of battery is normally distributed, with mean 128 hours and standard deviation 16 hours.
Let X be the life time of a battery.
when, X=100
the formula for z is:
z=X-mean/standard deviation
z= 100-128/16
z= -28/16
z=-1.75
P(X≤100)=P(z≤-1.75)
= P(z≥1.75)
=0.0409 (using z-dist table)
Hence, probability that a battery you buy lasts at most 100 hours is:
0.0409
Probability of an event is the measurement of chances of occurrence of that event. The probability that a battery bought will last at most 100 hours is [tex]P(X \leq 100) = 0.0401[/tex]
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]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]
Using the z scores will help to find the probabilities from the z tables(available online).
Let for the given context, the battery life in hours be tracked by a random variable X, then by the given data, we have:
[tex]X \sim N(128, 16)[/tex]
The needed probability is expressed as
[tex]P(X \leq 100)[/tex]
Converting the distribution to normal variate, we get:
[tex]Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 128}{16} \\\\Z \sim N(0,1)[/tex]
The needed probability converts to
[tex]P(X \leq 100) = P(Z < \dfrac{100 - 128}{16}) = p(Z < -1.75)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write [tex]P(Z < a) = P(Z \leq a)[/tex] )
Also, know that if we look for Z = a in z tables, the p value we get is [tex]P(Z \leq a) = p \: value[/tex]
Using the z tables, the p value for z score -1.75 is 0.0401
Thus,
[tex]P(X \leq 100) = P(Z < -1.75)= 0.0401[/tex]
Thus,
The probability that a battery bought will last at most 100 hours is given as
[tex]P(X \leq 100) = 0.0401[/tex]
Learn more about standard normal distribution here:
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