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a shipment of beach balls with a mean diameter of 28 cm and a standard deviation of 1.3 cm is normally distributed. By how many standard deviations does a beach ball with a diameter of 31.9 cm differ from the mean?
A.) 1
B.) 1.3
C.) 2
D.) 3

Verified Answer :

JeanaShupp

Answer:  D.) 3

Step-by-step explanation:

Given: Mean diameter: \(\mu=28\ cm\)

Standard deviation: \(\sigma=1.3\ cm\)

The required number of standard deviations is given by  :-

\(z=\frac{X-\mu}{\sigma}\\\\\Rightarrow\ z=\frac{31.9-28}{1.3}\\\\\Rightarrow\ z==3\)

Hence, a beach ball with a diameter of 31.9 cm is 3 standard deviations differ from the mean.

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