The probability that the mean of a sample of 78 computers would be less than 97.94 months exists 0.883.
A probability is a numerical representation of the likelihood or chance that a specific event will take place.
Let the given values be
[tex]\mu[/tex] = 99, [tex]$\sigma[/tex] = 8
Let x be the random variable that represents the life of a computer.
Sample size: n = 69
Then, the z-score corresponds to x = 97.94 will be :-
[tex]$z=\frac{97.94-99}{\frac{8}{\sqrt{69}} \quad[\because} \quad z=\frac{x-\mu}{\left.\frac{\sigma}{\sqrt{n}}\right]}$[/tex]
z = -1.10062
Required probability ( using standard z-value table ) :-
P(x < 97.94) = P(z < -1.10) = 1 - P(z < 1.10)
= 1 - 0.1170 = 0.883
Therefore, the probability that the mean of a sample of 78 computers would be less than 97.94 months exists 0.883.
To learn more about probability refer to:
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