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A certain chalkboard manufacturer determines that their largest blackboard model has a mean length of 5.00 m and a standard deviation of 1.0 cm. A certain school district orders 1000 of these chalkboards. How many are likely to have lengths of under 4.98 m?

Respuesta :

Answer:

23 chalkboards

Step-by-step explanation:

Given:

Mean length = 5 m

Standard deviation = 0.01

Number of units ordered = 1000

Now,

The z factor = [tex]\frac{\textup{x - Mean}}{\textup{standard deviation}}[/tex]

or

The z factor = [tex]\frac{\textup{4.98 - 5}}{\textup{0.01}}[/tex]

or

Z = - 2

Now, the Probability P( length < 4.98 )

Also, From z table the p-value = 0.0228

therefore,

P( length < 4.98 ) = 0.0228

Hence, out of 1000 chalkboard ordered (0.0228 × 1000) = 23 chalkboard are likely to have lengths of under 4.98 m.

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