Question

Experience has shown that the weight of oranges produced by an orchard in Tzaneen is normally distributed, with an average weight (μ) of 120g and a standard deviation (σ) of 20g. A sample of 115 of these oranges is randomly selected.

a) Determine the standard error for this sample. (3)

b) What is the probability that the mean weight of the oranges in the sample lies between 118g and 121.5g? Interpret your answer

Answer 1

Sol)

Given

Mean = 120

S.D = 20

n = 115

a)

Standard error = S.D / ✓n

S. E = 20 / ✓115

S.E = 1.86500

b)

Probability that sample lies between 118 and 121.5

P( 118< X < 121.5) = P ( ( X - mean ) / S.E < z < ( X - mean ) / S.E))

= P ( ( 118 - 120)/1.86500 < Z < ( 121.5-120)/1.86500))

= P ( -2 / 1.86500 < Z < 1.5/ 1.86500)

= P ( -1.0723860 < Z < 0.8042895)

= P( Z < 0.8042895) - P( Z < -1.0723860)

= 0.7894 - 0.1418

= 0.6476

= 64.76%

The probability that mean weight of oranges lies in the sample lies between 118 and 121.5 is 64.76%

That means 64.76% of samples lies between 118 and 121.5