Question

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

Answer 1

Answer:

The answer is "78.25%"

Step-by-step explanation:

Given:

$\mu =20\\\\\sigma=1.4$

$p(18.48$

                                 $=p(\frac{-1.52}{1.7}$

                                 $=p(Z$

                                 $=0.969258-0.1867329\\\\=0.7825251\\\\=78.25 \%$