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The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 47 and a standard deviation of 7. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 47 and 68

Sagot :

okpalawalter8

Answer:

[tex]P(-2<Z<2)=95\%[/tex]

Step-by-step explanation:

From question we are told that

Sample mean [tex]\=x= 47[/tex]

Standard deviation [tex]\sigma =7[/tex]

Generally the  X -Normal is given as

      [tex]Z=\frac{x-\=x}{\sigma}[/tex]

      [tex]Z=\frac{x-47}{9}[/tex]

Analyzing the range

    [tex]P(47<x<65) = P(0< z<2.00)[/tex]

    [tex]P(47<x<65) = 95/2[/tex]

   [tex]P(47<x<65) = 47.5\%[/tex]

Mathematically

[tex]Z_1 =\frac{47-47}{9} =0[/tex]

[tex]Z_2 =\frac{65-47}{9} =2[/tex]

Empirical rule shows that

[tex]P(-2<Z<2)=95\%[/tex]

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