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3 years ago

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A shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed. By how many

standard deviations does a ball bearing with a diameter of 58.2 mm differ from the mean?

2 answers:

asambeis [7]3 years ago

8 0

Mean 60 - 2(0.9) = 58.2
58.2 is 2 standard deviations from the mean.

tigry1 [53]3 years ago

8 0

Answer:

By 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.

Step-by-step explanation:

It is given that a shipment of racquetballs with a mean diameter of 60 mm and a standard deviation of 0.9 mm is normally distributed.

$Mean=60$

$\text{Standard deviation}=0.9$

Absolute difference written diameter of 58.2 mm and average diameter is

$|58.2-60|=1.8$

Divide the difference by standard deviation (i.e.,0.9), to find the by how many standard deviations does a ball bearing with a diameter of 58.2 mm differ from the mean.

$\frac{1.8}{0.9}=2$

Therefore 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.

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Step-by-step explanation:

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