According to the Government statistics, the average length of stay of Tourists v
ID: 3339608 • Letter: A
Question
According to the Government statistics, the average length of stay of Tourists visiting Melbourne was 2.5 days. Suppose the standard deviation of the length of stay for all tourists is 1 day. Assuming length of stay is normally distributed, answer the following (remember to show all workings and interpret your answer):
a. What is the probability that the length of stay will be between 1.5 and 3.5 days?
b. What is the average length of stay for the top 15% of tourists who stay the longest (assume a sample size of 100 was taken)?
c. Table 1 lists the average price and quantity consumed for a selection of food products. Calculate the Paasche price indexes using 2000 as the base year for years 2005 and 2010. Remember to interpret your answers
Year - Price (kg)
Year - Quantity (kg)
2000
2005
2010
2000
2005
2010
Sultanas
3.99
4.5
4.98
20
22
24
Bananas
2.99
3.98
13.49
13
11
10
Rice
12.85
13.99
18.99
10
15
12
Lentils
4.05
4.98
6.95
17
18
19
Table 1 lists the average price and quantity consumed for a selection of food products. Calculate the Paasche price indexes using 2000 as the base year for years 2005 and 2010. [3 marks]. Remember to interpret your answers
Year - Price (kg)
Year - Quantity (kg)
2000
2005
2010
2000
2005
2010
Sultanas
3.99
4.5
4.98
20
22
24
Bananas
2.99
3.98
13.49
13
11
10
Rice
12.85
13.99
18.99
10
15
12
Lentils
4.05
4.98
6.95
17
18
19
Explanation / Answer
A) P(1.5 < X < 3.5) = P((1.5 - 2.5)/1 < (X - mean)/SD < (3.5 - 2.5)/1)
= P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= 0.8413 - 0.1587
= 0.6826
B) P(X > x) = 0.15
Or, P((X - mean)/(SD / sqrt (n)) > (x - 2.5)/(1/sqrt (100)) = 0.15
Or, P(Z > (x - 2.5)/(1/10) = 0.15
Or, P(Z < (x - 2.5)/(1/10)) = 0.85
Or, (x - 2.5)/(1/10) = 1.04
Or, x = 2.5 + 1.04 * (1/10)
Or, x = 2.604
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