32% of adult internet users have purchased products or services online. For a random sampling of 200 adult users, find the mean, variance and standard deviation for the number who have purchased goods or services online?
Mean = np = (200)(.32) = ?
Variance = npq = (200)(.32)(.68) = ?
(Note: q = 1 - p)
Standard deviation = square root of the variance
I'll let you finish the calculations.
43.68
Oh, I see you're trying to calculate some statistics. Well, let's dive into the world of online shopping statistics!
Given that 32% of adult internet users have purchased products or services online, we can use this information to find the mean, variance, and standard deviation for a random sampling of 200 adult users.
To find the mean, we multiply the sample size (200) by the percentage (32%) to get the expected number of people who have purchased goods or services online. So, the mean is:
Mean = Sample Size × Percentage
= 200 × 0.32
= 64
Great! Now, let's move on to the variance. The formula for variance is:
Variance = Sample Size × Percentage × (1 - Percentage)
Variance = 200 × 0.32 × (1 - 0.32)
= 200 × 0.32 × 0.68
= 43.52
Lastly, we can calculate the standard deviation by taking the square root of the variance. Therefore:
Standard Deviation = √Variance
= √43.52
≈ 6.60
So, for a random sampling of 200 adult internet users, the mean number of individuals who have purchased goods or services online would be 64, the variance would be 43.52, and the standard deviation would be approximately 6.60. Keep in mind that these are just statistical estimates. Enjoy shopping and stay safe online!
To find the mean, variance, and standard deviation for the number of adult internet users who have purchased products or services online, we can use the formula for a binomial distribution.
The mean (μ) for a binomial distribution is given by the formula μ = n * p, where n is the number of trials and p is the probability of success in each trial.
In this case, n = 200 (the number of adult users) and p = 32% (or 0.32, the probability of an adult user purchasing goods or services online).
So, the mean μ = 200 * 0.32 = 64.
The variance (σ^2) for a binomial distribution is given by the formula σ^2 = n * p * (1 - p).
In this case, σ^2 = 200 * 0.32 * (1 - 0.32) = 43.52.
Finally, the standard deviation (σ) is the square root of the variance, so σ = √(43.52) = 6.60 (approximately).
Therefore, the mean is 64, the variance is 43.52, and the standard deviation is 6.60 for the number of adult internet users who have purchased products or services online in a random sample of 200 users.
To find the mean, variance, and standard deviation, we need to use the formulas and the given data.
Given:
Percentage of adult internet users who have purchased products/services online = 32%
Total adult internet users (sample size) = 200
Step 1: Calculate the mean (expected value)
The mean is found by multiplying the sample size by the percentage and dividing by 100.
Mean = (Sample size * Percentage) / 100
Mean = (200 * 32) / 100
Mean = 64
Therefore, the mean number of adult users who have purchased goods or services online is 64.
Step 2: Calculate the variance
The variance is a measure of how spread out the numbers are from the mean. To calculate the variance, we need to subtract the mean from each value, square the result, and then sum up all the squared differences. Finally, divide the sum by the sample size.
Variance = Sum of [(Value - Mean)^2] / Sample size
Since we don't have the actual values, we can work with percentages instead. We'll calculate the variance based on the percentage instead of absolute numbers.
Percentage Variance = Sum of [(Percentage - Mean)^2] / Sample size
Percentage Variance = [(32 - 32)^2 + (68 - 32)^2] / 200
Percentage Variance = (36^2 + 68^2) / 200
Percentage Variance = 1296 + 4624 / 200
Percentage Variance = 5920 / 200
Percentage Variance = 29.6
Therefore, the variance for the number of users who have purchased goods or services online is 29.6.
Step 3: Calculate the standard deviation
The standard deviation is the square root of the variance.
Standard Deviation = Square Root of Variance
Standard Deviation = Square Root of 29.6
Standard Deviation ≈ 5.44
Therefore, the standard deviation for the number of users who have purchased goods or services online is approximately 5.44.
To summarize:
Mean: 64
Variance: 29.6
Standard Deviation: 5.44