Question

An orange juice producer buys oranges from a large orange grove that has one variety of orange. The amount of juice squeezed from these oranges is approximately normally​ distributed, with a mean of 4.40 ounces and a standard deviation of 0.32 ounce. Suppose that you select a sample of 16 oranges.

a. What is the probability that the sample mean amount of juice will be at least 4.27 ​ounces?

b. The probability is 72​% that the sample mean amount of juice will be contained between what two values symmetrically distributed around the population​ mean?

c. The probability is 74​% that the sample mean amount of juice will be greater than what​ value?

Answer 1

Solution :

Given that,

mean = = 4.40

standard deviation = = 0.32

n = 16

= = 4.40

= / n = 0.32 / 16 = 0.08

a) P( 4.27) = 1 - P( 4.27)

= 1 - P[( - ) / (4.27 - 4.40) / 0.08 ]

= 1 - P(z -1.625)

Using z table,    

= 1 - 0.0521

= 0.9479

b) Using standard normal table,

P( -z < Z < z) = 72%

= P(Z < z) - P(Z <-z ) = 0.72

= 2P(Z < z) - 1 = 0.72

= 2P(Z < z) = 1 + 0.28

= P(Z < z) = 1.28 / 2

= P(Z < z) = 0.64

= P(Z < 0.358) = 0.64

= z  ± 0.358

Using z-score formula  

= z * +

= -0.358 * 0.08 + 4.40

= 4.37 ounce.

Using z-score formula  

= z * +

= 0.358 * 0.08 + 4.40

= 4.43 ounce.

two values is between 4.37 and 4.43 ounce.

c) Using standard normal table,

P(Z > z) = 74%

= 1 - P(Z < z) = 0.74  

= P(Z < z ) = 1 - 0.74

= P(Z < z ) = 0.26

= P(Z < -0.643 ) = 0.26  

z = -0.643

Using z-score formula  

= z * +

= -0.643 * 0.08 + 4.40

= 4.35 ounce.