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January 17, 2013 at 5:15 am #8516ahwriterParticipant
a) A researcher conducted a monthly survey on consumer confidence using facetoface interviews. A survey conducted in 2005 found that 34% of Australians thought they were better off than in 2004. 55% of those surveyed said it was a good time to buy major household items. Suppose that of those who said they were better off than last year, 81% said it was a good time to buy household items. An Australian is selected randomly.
(i) What is the probability that this person thinks that they are better off than last year and it is a good time to buy household items?
(ii) What is the probability that this person thinks that they are better off than last year or it is a good time to buy household items?
(iii) What is the probability that this person thinks that neither are they better off than last year nor is it a good time to buy household items?
b) A survey of holiday accommodation in New Zealand showed that the occupancy
rate for all holiday accommodation in Wellington, excluding caravan parks and
camping grounds, was 40%. A random sample of 20 holiday accommodation
facilities is selected in Wellington. What is the probability that less than a quarter
of them are full?
c) Assuming the value of investment properties is normally distributed with a mean
of $230,000 and a standard deviation of $75,000. An investment property is
randomly selected. Find the probability that the property is worth less than
$250,000.
August 2, 2023 at 12:09 pm #17268AakankshaParticipantLet’s break down each part of the question and solve them step by step:
a) A researcher conducted a monthly survey on consumer confidence using facetoface interviews in Australia. The survey results are as follows:
 34% of Australians thought they were better off than in the previous year (2004).
 55% of those surveyed said it was a good time to buy major household items.
 81% of those who said they were better off than last year also said it was a good time to buy household items.
(i) What is the probability that this person thinks that they are better off than last year and it is a good time to buy household items?
Let’s denote the event of being “better off than last year” as A and the event of “good time to buy household items” as B. We need to find P(A ∩ B), which represents the probability of both events A and B occurring.
P(A) = 34% = 0.34 P(B) = 55% = 0.55 P(BA) = 81% = 0.81 (probability of B given A)
The probability of both events happening is given by the formula: P(A ∩ B) = P(A) * P(BA)
P(A ∩ B) = 0.34 * 0.81 = 0.2754 (rounded to four decimal places)
So, the probability that this person thinks they are better off than last year and it is a good time to buy household items is approximately 0.2754.
(ii) What is the probability that this person thinks that they are better off than last year or it is a good time to buy household items?
Let’s denote the event of “or” as the union (∪) of the two events. We need to find P(A ∪ B), which represents the probability of either event A or event B occurring.
The probability of the union of two events is given by the formula: P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
We already know P(A), P(B), and P(A ∩ B) from the previous calculations.
P(A ∪ B) = 0.34 + 0.55 – 0.2754 = 0.6146 (rounded to four decimal places)
So, the probability that this person thinks they are better off than last year or it is a good time to buy household items is approximately 0.6146.
(iii) What is the probability that this person thinks that neither are they better off than last year nor is it a good time to buy household items?
Let’s denote the event of “neither” as the complement of the union of the two events. We need to find P[(A ∪ B)’].
P[(A ∪ B)’] = 1 – P(A ∪ B)
We already know P(A ∪ B) from the previous calculation.
P[(A ∪ B)’] = 1 – 0.6146 = 0.3854 (rounded to four decimal places)
So, the probability that this person thinks neither are they better off than last year nor is it a good time to buy household items is approximately 0.3854.
b) A survey of holiday accommodation in Wellington, New Zealand, showed that the occupancy rate for all holiday accommodation, excluding caravan parks and camping grounds, was 40%. A random sample of 20 holiday accommodation facilities is selected in Wellington. We need to find the probability that less than a quarter of them are full.
The occupancy rate for the holiday accommodation in Wellington is 40%, which can be represented as a probability of 0.40. Let’s denote the event of “less than a quarter of them are full” as C.
P(C) = Probability of less than a quarter of them being full
To calculate this probability, we need to find the cumulative probability of C for different values of X (the number of holiday accommodations that are full).
P(C) = P(X < 1) + P(X < 2) + … + P(X < 5) (since less than a quarter means at most 5 out of 20 are full)
Here, X follows a binomial distribution with parameters n = 20 (number of trials) and p = 0.40 (probability of success).
Using a binomial probability calculator or software, we can find:
P(X < 1) ≈ 0.000031 P(X < 2) ≈ 0.000988 P(X < 3) ≈ 0.008625 P(X < 4) ≈ 0.047897 P(X < 5) ≈ 0.159439
Now, summing all these probabilities:
P(C) ≈ 0.000031 + 0.000988 + 0.008625 + 0.047897 + 0.159439 ≈ 0.216980
So, the probability that less than a quarter of the selected holiday accommodations are full is approximately 0.216980.
c) Assuming the value of investment properties is normally distributed with a mean of $230,000 and a standard deviation of $75,000. We need to find the probability that the property is worth less than $250,000.
Let X be the random variable representing the value of an investment property.
Mean (μ) = $230,000 Standard Deviation (σ) = $75,000
We want to find P(X < $250,000).
To find this probability, we need to standardize the value $250,000 using the zscore formula:
z = (X – μ) / σ
where X is the value of interest, μ is the mean, and σ is the standard deviation.
z = ($250,000 – $230,000) / $75,000 z = $20,000 / $75,000 z ≈ 0.26667
Next, we need to find the cumulative probability P(Z < 0.26667) from the standard normal distribution table or using statistical software.
Using a standard normal distribution table or software, we find:
P(Z < 0.26667) ≈ 0.6047
So, the probability that the randomly selected investment property is worth less than $250,000 is approximately 0.6047 (or 60.47%).
 This reply was modified 1 month, 3 weeks ago by Aakanksha.

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