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

The
weight of oranges are normally distributed with a mean weight of
150 grams and a standard deviation of 10 grams. in a sample of 100
oranges, how many will weigh between 130 and 170 grams?

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