Question

The weight of navel oranges of a domestic farm is normally distributed with a mean of 8.0oz and a standard deviation of 1.5oz. Suppose that you bought 10 oranges randomly sampled.

a) what are the mean of the sampling distribution and the standard error of the mean?

b) what is the probability that the sample mean is between 8.5 and 10.0 oz?

c) The probability is 90% that the sample mean will be between what two values symmetrically distributed around the population mean?

Answer 1

Solution :

Given that,

mean = = 8.0

standard deviation = = 1.5

n = 10

a) =   = 8.0

= / n = 1.5 / 10 = 0.474

b) P(8.5 < < 10.0)  

= P[(8.5 - 8.0) /0.474 < ( - ) / < (10.0 - 8.0) / 0.474 )]

= P(1.05 < Z < 4.22)

= P(Z < 4.22) - P(Z < 1.05)

Using z table,  

= 1 - 0.8531

= 0.1469

c) Using standard normal table,

P( -z < Z < z) = 90%

= P(Z < z) - P(Z <-z ) = 0.90

= 2P(Z < z) - 1 = 0.90

= 2P(Z < z) = 1 + 0.90

= P(Z < z) = 1.90 / 2

= P(Z < z) = 0.85

= P(Z < 1.04) = 0.85

= z  ± 1.04

Using z-score formula,

x = z * +

x = -1.04 * 0.474 + 8.0

x = 7.5

Using z-score formula,

x = z * +

x = 1.04 * 0.474 + 8.0

x = 8.5

90% two values =( 7.5, 8.5)