3. According to a survey, only 15% of customers who visited the web site of a major retail store made a purchase. Random samples of size 50 are selected.
3.1 Find the mean of all the sample proportions of customers who will make a purchase after visiting the web site.
3.2 Find the standard deviation of all the sample proportions of customers who will make a purchase after visiting the web site.
3.3 What proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the web site?
3.4 What proportion of the samples will have less than 15% of customers who will make a purchase after visiting the web site?
3.5 What is the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the web site ?
3.6 If 90% of the samples will have less than, what percentage of customers who will make a purchase after visiting the web site? [ P(p ̂<k)=0.90 find k ]
3.1 To find the mean of all the sample proportions of customers who will make a purchase after visiting the website, we use the formula for the mean of a sampling distribution of sample proportions:
Mean = p = 0.15
So, the mean of the sample proportions is 0.15.
3.2 To find the standard deviation of all the sample proportions of customers who will make a purchase after visiting the website, we use the formula for the standard deviation of a sampling distribution of sample proportions:
Standard Deviation = sqrt( (p * (1-p)) / n )
where p = 0.15 and n = 50
Standard Deviation = sqrt( (0.15 * (1-0.15)) / 50 )
Standard Deviation = sqrt( 0.1275 / 50 )
Standard Deviation = sqrt(0.00255)
Standard Deviation = 0.0505
So, the standard deviation of the sample proportions is 0.0505.
3.3 To find the proportion of the samples that will have between 20% and 30% of customers who will make a purchase after visiting the website, we need to find the z-scores for both 20% and 30% and then find the area between these two z-scores in the standard normal distribution.
Z-score for 20%: (0.2 - 0.15) / sqrt( (0.15 * (1-0.15)) / 50 ) = 0.05 / 0.0505 = 0.9901
Z-score for 30%: (0.3 - 0.15) / sqrt( (0.15 * (1-0.15)) / 50 ) = 0.15 / 0.0505 = 2.9703
Using a standard normal distribution table or calculator, we find the area to the left of z = 0.9901 is 0.8383 and the area to the left of z = 2.9703 is 0.9984. Therefore, the area between these two z-scores is 0.9984 - 0.8383 = 0.1601.
So, the proportion of samples that will have between 20% and 30% of customers who will make a purchase after visiting the website is 0.1601.
3.4 To find the proportion of the samples that will have less than 15% of customers who will make a purchase after visiting the website, we need to find the z-score for 15% and then find the area to the left of this z-score in the standard normal distribution.
Z-score for 15%: (0.15 - 0.15) / sqrt( (0.15 * (1-0.15)) / 50 ) = 0 / 0.0505 = 0
The area to the left of z = 0 is 0.5.
Therefore, the proportion of samples that will have less than 15% of customers who will make a purchase after visiting the website is 0.5.
3.5 To find the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the website, we need to find the z-score for 30% and then find the area to the right of this z-score in the standard normal distribution.
Z-score for 30%: (0.3 - 0.15) / sqrt( (0.15 * (1-0.15)) / 50 ) = 0.15 / 0.0505 = 2.9703
The area to the right of z = 2.9703 is 1 - 0.9984 = 0.0016.
So, the probability that a random sample of 50 will have at least 30% of customers who will make a purchase after visiting the website is 0.0016.
3.6 To find the percentage of customers who will make a purchase after visiting the website such that 90% of the samples will have less than this proportion, we need to find the z-score corresponding to the 90th percentile of the standard normal distribution.
Since we want 90% of the samples to have less than this proportion, we are looking for the z-score that corresponds to the cumulative probability of 0.90 (or 90th percentile).
From the standard normal distribution table or using a calculator, we find the z-score corresponding to a cumulative probability of 0.90 is approximately 1.28.
Now, we solve for the proportion:
Z = (k - 0.15) / 0.0505
1.28 = (k - 0.15) / 0.0505
1.28 * 0.0505 = k - 0.15
k = (1.28 * 0.0505) + 0.15
k = 0.1858
Therefore, 90% of the samples will have less than approximately 18.58% of customers who will make a purchase after visiting the website.