"A sample of 20 randomly chosen water melons was taken from a large population, and their weights were measured. The mean weight of the sample was 105 lb. and the standard deviation was 15 lb. Calculate (correct to one decimal place) 99.5% confidence limits for the mean weight of the whole population of watermelons."

Respuesta :

Following are the calculation to the value:

Given:

[tex]n= 20\\\\\mu= 105\\\\\sigma= 15\\\\CI= 99.5\%\\\\[/tex]

To find:

value=?

Solution:

[tex]\to z= 2.807 for 99.5\%\\\\\to 2.807=\frac{x-105}{\frac{15}{\sqrt{20}}}\\\\\to 2.807 \times\frac{15}{\sqrt{20}} =x-105\\\\\to 2.807 \times\frac{15}{\sqrt{20}}+105 =x\\\\\to x= \to 2.807 \times\frac{15}{\sqrt{20}}+105\\\\[/tex]

Solving the value for x\\\\

[tex]\to x= 114.41\\\\[/tex]

Therefore, the final answer is "114.41".

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