Answer:
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 95
Given that the standard deviation of the Population = 5
Let 'X' be the random variable in a normal distribution
Let X⁻ = 96.3
Given that the size 'n' = 84 monitors
<u><em>Step(ii):-</em></u>
<u><em>The Empirical rule</em></u>


Z = 2.383
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = P(Z≥2.383)
= 1- P( Z<2.383)
= 1-( 0.5 -+A(2.38))
= 0.5 - A(2.38)
= 0.5 -0.4913
= 0.0087
<u><em>Final answer:-</em></u>
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Answer:
O x = 2
Step-by-step explanation:
Only x-intercepts will make the line horizontally and x = 2 is the line that passes through the point (2, 10)
Answer is 49
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<u>Answer:</u>
cubic feet
<u>Step-by-step explanation:</u>
We are given that the weight of water is 62 and one half lb per cubic foot and we are to find the number of cubic feet that would be occupied by 180 lb of water.
For that, we will divide total amount of water by per unit weight of water:
Cubic feet occupied by 180 lb of water =
= 2.88 cubic feet or
cubic feet