Question

An orange juice producer buys all his oranges from a large orange orchard. The amount of juice squeezed from the oranges is approximately normally distributed with a mean of 4.70 ounces and a standard deviationof 0.40 ounces a.(10) What is the probability that a randomly selected orange will contain i. Between 4.70 and 5.00 ounces? ii. Between 5.00 and 5.50 ounces? ii. Between 4.00 and 5.00 ounces iv. More than 4.00 ounces b. (5) Seventy-seven percent of the oranges will contain at least how many ounces of juice (Suppose that a sample of 25 oranges are selected) c.(5) What is the probability that the sample mean will be at least 4.60 ounces? d. (5) There is a 70 percent chance that the sample average will fall between which two values symmetrically distributed around the population mean

Answer 1

Solutiona:i:

P(4.70<X<5.00)

P(4.70-4.70/0.40<Z<5-4.7/0.4)

=P(0<Z<0.75)

=P(Z<0.75)-P(Z<0)

=0.7734-0.5

=0.2734

Soluton1b:

P(5-4.70/0.40<Z<5.5-4.7/0.4)

P(.75<Z<2)

P(Z<2)-P(Z<0.75)

=0.9772-0.7734

=0.2038

Solution1c:

P(4<x<5)

P(4-4.7/0.40<Z<5-4.7/0.40)

=p(-1.75<z<0.75)

=P(Z<0.75)-P(Z<-1.75)

=0.7333

Solutin1d:

P(X>4)

P(z>4-4.7/0.4)

P(Z>-1.75)

=P(Z<1.75)

=0.9599

Normal Curve, mean 4.7 , SD 0.4 Normal Curve, mean 4.7, SD- 0.4 Shaded Area 0.2039 Shaded Area 0.2734 4.7 5 5.5 Normal Curve, mean Shaded Area 4.7 , SD = 0.4 0.7333 Normal Curve, mean 4.7 , SD = 0.4 Shaded Area 0.9599