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 6.0 ounces and a standard deviation of 0.48 ounce. Suppose that you select a sample of 36 oranges.

a. What is the probability that the sample mean amount of juice will be at least 5.36 ounces?

b. The probability is 70​% 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 ,

$\mu$
_T
= 6.0

$\sigma$
_T
= $\sigma$ / $\sqrt$ n = 0.48 / $\sqrt$ 36 = 0.08

a) P(
T
$\geq$ 5.36) = 1 - P(
T
$\leq$ 5.36)

= 1 - P[(
T
- $\mu$
_T
) / $\sigma$
_T
$\leq$ (5.36 - 6.0) / 0.08]

= 1 - P(z $\leq$ -8.00)   

= 1 - 0

= 1

b) Using standard normal table,

P( -z < Z < z) = 70%

= P(Z < z) - P(Z <-z ) = 0.70

= 2P(Z < z) - 1 = 0.70

= 2P(Z < z) = 1 + 0.70

= P(Z < z) = 1.70 / 2

= P(Z < z) = 0.85

= P(Z < 1.036) = 0.85

= z  ± 1.036

Using z-score formula,

T
= z * $\sigma$
_T
+ $\mu$
_T

= -1.036 * 0.08 + 6.0

T

= 5.92 ounces

T

Using z-score formula,

T
= z * $\sigma$
_T
+ $\mu$
_T

= 1.036 * 0.08 + 6.0

T

= 6.08 ounces

T

The probability is 70​% that the sample mean amount of juice will be contained between 5.92 ounces and 6.08 ounces.

c) Using standard normal table,

P(Z > z) = 74%

= 1 - P(Z < z) = 0.74  

= P(Z < z ) = 1 - 0.74

= P(Z < -0.643 ) = 0.26  

z = -0.643

Using z-score formula  

T
= z * $\sigma$
_T
+ $\mu$
_T

= -0.643 * 0.08 + 6.0

T

= 5.95 ounces.

T

The probability is 74​% that the sample mean amount of juice will be greater than 5.95 ounces.