In one online store, $12 \%$ of the orders are placed by new customers. Let $N$ be the number of orders placed in a day to reach the store's first new customer. Assume each order's customer status is independent.

Find the mean and standard deviation of $N$.
Round your answers to one decimal place.
Mean: $\square$ orders

Standard deviation: $\square$ orders

Question

In one online store, $12 \%$ of the orders are placed by new customers. Let $N$ be the number of orders placed in a day to reach the store's first new customer. Assume each order's customer status is independent.

Find the mean and standard deviation of $N$.
Round your answers to one decimal place.
Mean: $\square$ orders

Standard deviation: $\square$ orders

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the mean of the geometric distribution with probability $p = 0.12$, we use the formula $\mu = \frac{1}{p}$. ‖ ‖ step 2 ⋮ Substituting $p = 0.12$ into the formula gives us $\mu = \frac{1}{0.12}$. ‖ ‖ step 3 ⋮ Calculating the mean, we get $\mu = \frac{1}{0.12} = 8.33333$. ‖ ‖ step 4 ⋮ To find the standard deviation of the geometric distribution, we use the formula $\sigma = \frac{\sqrt{1-p}}{p}$. ‖ ‖ step 5 ⋮ Substituting $p = 0.12$ into the formula gives us $\sigma = \frac{\sqrt{1-0.12}}{0.12}$. ‖ ‖ step 6 ⋮ Calculating the standard deviation, we get $\sigma = \frac{\sqrt{0.88}}{0.12} = 7.81736$. ‖ ‖ step 7 ⋮ Round the mean and standard deviation to one decimal place: Mean = 8.3 orders, Standard deviation = 7.8 orders. ‖ ∻1 Answer∻ ⚹ Mean: 8.3 orders ⚹ ⚹ Standard deviation: 7.8 orders ⚹ ∻Key Concept∻ ⚹Geometric Distribution⚹ ∻Explanation∻ ⚹The mean of a geometric distribution with probability $p$ is given by $\frac{1}{p}$, and the standard deviation is given by $\frac{\sqrt{1-p}}{p}$. These formulas are used to calculate the expected number of trials until the first success and the variability of this number, respectively.⚹◊What is the formula to calculate the mean and standard deviation of N, given that 12% of the orders are placed by new customers?⍭ Generate me a similar question◊

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