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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
Before we get started, it is good to remember PEMDAS - the acronym that tells you the order to carry out equations.
P = parentheses
E = exponents
M = multiplication
D = division
A = addition
S = subtraction
Knowing this, we should start by doing the equations <em>within </em>each parentheses.
So we did the addition and subtraction pieces within each group leaving us with the above equation. Now let's multiply:
20 is the answer.
P = parentheses
E = exponents
M = multiplication
D = division
A = addition
S = subtraction
Knowing this, we should start by doing the equations <em>within </em>each parentheses.
So we did the addition and subtraction pieces within each group leaving us with the above equation. Now let's multiply:
20 is the answer.