- 20 Marks
Question
a) Receipts at a particular depot have amounts which follow the Normal distribution with a mean of GH¢103.60 and a standard deviation of GH¢8.75.
Required:
i) Determine the percentage of receipts over GH¢120.05.
ii) Determine the percentage of receipts below GH¢92.75.
iii) Determine the percentage of receipts between GH¢83.65 and GH¢117.60.
iv) Determine the receipts amount such that approximately 25 percent of receipts are greater.
v) Above what amount will 90 percent of receipts lie?
b) If 10.56 percent of receipts have an amount above GH¢110.05 and 4.01 percent of receipts have an amount above GH¢120.05.
Required:
Calculate the mean and standard deviation of the receipts assuming that they are normally distributed.
Answer
a)
i)
From Z-tables: Pr (Z > 1.88) = 0.0301
Thus, approximately 3% of the receipts are over GH¢120.05.
ii)
Pr (X < 92.75) = 
From Z-tables:
Pr (Z < −1.24) = 0.1075
Thus, approximately 10.75% of the receipts are below GH¢92.75.
iii)
Pr(83.65 < X < 117.60) = Pr 
From Z-tables:
Pr (−2.28 < Z < 1.60) = 0.9432
Thus, approximately 94.32% of the receipts lie between GH¢83.65 and GH¢117.60.
iv)
To find the value of k such that 
we use the Z-value for the 75th percentile (since 25% of receipts are above):
Pr (Z = 0.67)
Thus,
k = 103.60 + 0.67 × 8.75 = 109.102
Therefore, the amount is GH¢109.10.
v)
To find the value of kk such that 90% of receipts are below: Pr (Z = 1.28)
Thus,
k = 103.60 + 1.28 × 8.75 = 114.80
Thus, GH¢114.80 is the amount below which 90% of receipts lie.
b)
Solve equation (1) and two simultaneously for the parameters as follows
Subtract equation (1) from (2) 
Substitute 
in the equation (10) we have
- Tags: Mean, Normal distribution, Probability, Receipts, Standard Deviation, Z-scores
- Level: Level 1
- Topic: Probability
- Series: MAY 2016
- Uploader: Kwame Aikins