The weight of oranges growing in an orchard is normally distributed with a mean weight of 7 oz. and a standard deviation of 1.5 oz. Using the empirical rule, determine what interval would represent weights of the middle 99.7% of all oranges from this orchard.

The percentage is 95.45%

Step-by-step explanation:

In this question, we are tasked with calculating the percentage of oranges weighing between a particular range.

To do this , we use the z-scores of both numbers in the range.

Mathematically, z-score can be calculated as (x - Mean)/SD

Thus we have P(3 <x < 7) = P( (3-5)/1 < z < (7-5)/1)

= P( -2 < z < 2)

Using standard score table, we have

P(-2<z<2) = 0.9545

The percentage is thus 9545/100 = 95.45

54

step-by-step explanation:

step-by-step explanation:

given perimeter: 28cm

length: 9cm

atp,

2(l+w)=p of the reactangle

2(9+x)=28

18+2x=28

2x=28-18

x=10/2

x=5

therefore width is 5 cm