contestada

assume that the weight of ripe watermelons grown at a particular farm are normally distributed witha mean of 20 pounds and a standard deviation of 1.4 pounds. determine the percent of watermelons that weigh between 18.48 pounds and 22.68 pounds

Respuesta :

Answer:

The answer is "78.25%"

Step-by-step explanation:

Given:

[tex]\mu =20\\\\\sigma=1.4[/tex]

[tex]p(18.48<x<22.63)=p(\frac{18.48-20}{1.7}<\frac{\bar{x}-\mu}{\sigma}<\frac{22.63-20}{1.4})\\\\[/tex]

                                 [tex]=p(\frac{-1.52}{1.7}<Z<\frac{2.63}{1.4})\\\\=p(-0.89<Z<1.87)\\\\[/tex]

                                 [tex]=p(Z<1.87)-p(Z<-0.89)\\\\[/tex]

                                 [tex]=0.969258-0.1867329\\\\=0.7825251\\\\=78.25 \%[/tex]

Otras preguntas