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 racquetball with a diameter of 58.2 mm differ from the mean?

0.9
1
2
3

Respuesta :

Answer:

C- 2

Step-by-step explanation:

60 - 0.9 X 2 = 58.2

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