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Answer:
<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>
<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>
<u>Step-by-step explanation</u>:
<u><em>Explanation</em></u>:-
Given mean of the Population 'μ ' = 51 months
Standard deviation of the Population 'σ' = 11 months
Let 'X' be the random variable of Normal distribution
<em>Let 'X' = 73</em>
<em></em><em></em>
<em>Let 'X' = 84</em>
<em></em><em></em>
<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>
<em>P( 73 < x < 84 ) = P( 2 < Z < 3)</em>
<em> = P( Z<3) - P( Z <2)</em>
<em> = 0.5 + A(3) - ( 0.5 + A(2))</em>
<em> = A(3) - A( 2)</em>
<em> = 0.49865 - 0.4772 ( From Normal table)</em>
<em> = 0.02145</em>
<em> P( 73 < x < 84 ) = 0.02145</em>
<em>The approximate percentage of cars that remain in service between 73 and 84 months</em>
<em>P( 73 < x < 84 ) = 0.02145 = 2.1 %</em>
Final Answer:
Steps/Reasons/Explanation:
Question: Solve for .
.
= ?
<u>Step 1</u>: Divide both sides by .
<u>Step 2</u>: Simplify to
.
~I hope I helped you :)~