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Sagot :
Answer:
52.38%
Step-by-step explanation:
P(991<X<997) = normalcdf(991,997,993,4) = 0.5328, therefore, about 52.38% of the bottles have volumes between 991 mL and 997 mL
The required proportion of bottles have volumes between 991 mL and 997 mL is 53.28%
Given that,
A bottler of drinking water fills plastic bottles with a mean volume of 993 milliliters (mL),
And standard deviation of 4 ml. The fill volumes are normally distributed.
We have to determine,
What proportion of bottles have volumes between 991 mL and 997 mL.
According to the question,
Mean of the drinking bottle = 993ml
[tex]\mu = 993[/tex]
And standard deviation of = 4 mL
[tex]\sigma = 4[/tex]
The proportion of bottles have volumes between 991 mL and 997 mL is,
[tex]= P (991<X<997) \\\\= P ( \dfrac{991-993}{4} <X < \dfrac{997-993}{4}) \\\\P( -0.5 < X < 1) = 0.5328 = 53.28\ percent[/tex]
Hence, The required proportion of bottles have volumes between 991 mL and 997 mL is 53.28%.
To know more about Proportion click the link given below.
https://brainly.com/question/4312900
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