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.
[tex]Mean=60[/tex]
[tex]\text{Standard deviation}=0.9[/tex]
Absolute difference written diameter of 58.2 mm and average diameter is
[tex]|58.2-60|=1.8[/tex]
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.
[tex]\frac{1.8}{0.9}=2[/tex]
Therefore 2 standard deviations a ball does bearing with a diameter of 58.2 mm differ from the mean.
