Respuesta :
Answer:
ans=13.59%
Step-by-step explanation:
The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], we have these following probabilities
[tex]Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827 [/tex]
[tex]Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545[/tex]
[tex]Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973[/tex]
In our problem, we have that:
The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months
So [tex]\mu = 53, \sigma = 11[/tex]
So:
[tex]Pr(53-11 \leq X \leq 53+11) = 0.6827[/tex]
[tex]Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545[/tex]
[tex]Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973[/tex]
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[tex]Pr(42 \leq X \leq 64) = 0.6827[/tex]
[tex]Pr(31 \leq X \leq 75) = 0.9545[/tex]
[tex]Pr(20 \leq X \leq 86) = 0.9973[/tex]
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What is the approximate percentage of cars that remain in service between 64 and 75 months?
Between 64 and 75 minutes is between one and two standard deviations above the mean.
We have [tex]Pr(31 \leq X \leq 75) = 0.9545 = 0.9545[/tex] subtracted by [tex]Pr(42 \leq X \leq 64) = 0.6827[/tex] is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.
To find just the percentage above the mean, we divide this value by 2
So:
[tex]P = {0.9545 - 0.6827}{2} = 0.1359[/tex]
The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.
The mean and standard deviation of 53 and 11, using the 68-95-99.7 rule, gives the percentage of cars remaining between 64 and 75 months as 13.59%
How can the percentage of cars in service be found?
The mean number of months in service = 53 months
The standard deviation = 11 months
The 68-95-99.7 rule is presented as follows;
[tex]pr( \mu - \sigma \: \leqslant \mu \leqslant \mu + \sigma) = 68.27\%[/tex]
[tex]pr( \mu - 2\sigma \: \leqslant \mu \leqslant \mu + 2\sigma) = 95.45\%[/tex]
[tex]pr( \mu - 3\sigma \: \leqslant \mu \leqslant \mu + 3\sigma) = 99.73\%[/tex]
Which gives;
P(53-11<= Mean <= 53 + 11) = 68.27%
P(42 <= Mean <= 64) = 68.27%
P(31 <= Mean <= 75) = 95.45%
P(20 <= Mean <= 86) = 99.73%
Between 64 and 75 months, we have;
0.9545 - 0.68.27 = P((42 - 31) + (75 - 64))
Therefore;
P(75 - 64) = (0.9545 - 0.6827)÷2= 0.1359 = 13.59%
The percentage of cars that will remain in service between 64 and 75 months is approximately 13.59%
Learn more about probability distributions here:
https://brainly.com/question/11290242
