Question

1. amazon prime is a subscription service for the online retailer amazon.com that gives the subscriber various perks such as free two - day shipping on most of the items sold by amazon. according to capitol one shopping, amazon prime members purchase a mean of 6.4 items per month, with a standard deviation of 2.3 items. assume the distribution of items purchased per month is approximately normal.

a. what is the probability that a randomly selected amazon prime member purchases at least ten items per month?
b. a researcher for amazon obtains the purchasing records for a random sample of 100 amazon prime members. he is interested in determining the number of amazon prime members that order at least ten items per month. define the random variable of interest, t.
c. could the random variable defined in part b be considered a binomial random variable? explain.
d. find the probability that exactly five amazon prime members in the sample purchase at least ten items per month?

Explanation:

Step 1: Identify the distribution and parameters

We know that the number of items purchased per month for an individual member follows a normal distribution with mean $\mu = 6.4$ and standard deviation $\sigma=2.3$. We want to find $P(X\geq10)$, where $X$ is the number of items purchased per month. First, we calculate the z - score using the formula $z=\frac{x - \mu}{\sigma}$.

For $x = 10$, the z - score is $z=\frac{10 - 6.4}{2.3}=\frac{3.6}{2.3}\approx1.57$

Step 2: Find the probability using the z - table

We know that $P(X\geq10)=1 - P(X\lt10)$. Since $X$ is normally distributed, $P(X\lt10)=\Phi(z)$ (where $\Phi(z)$ is the cumulative distribution function of the standard normal distribution). Looking up $z = 1.57$ in the standard normal table, we find that $\Phi(1.57)\approx0.9418$.

So $P(X\geq10)=1 - 0.9418 = 0.0582\approx0.058$ (or approximately 0.059 as in the hand - written note, due to possible differences in z - table precision or rounding during z - score calculation)

For part (d), if we assume that the number of members who purchase at least 10 items follows a binomial distribution with $n = 100$ (number of trials), $p=P(X\geq10)\approx0.058$ (or 0.059 as in the note) and $x = 5$ (number of successes). The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$

$C(100,5)=\frac{100!}{5!(100 - 5)!}=\frac{100\times99\times98\times97\times96}{5\times4\times3\times2\times1}=75287520$

$p = 0.059$, $1-p = 0.941$, $k = 5$, $n = 100$

$P(X = 5)=75287520\times(0.059)^{5}\times(0.941)^{95}$

First, $(0.059)^{5}\approx0.059\times0.059\times0.059\times0.059\times0.059\approx0.059^{5}\approx7.14\times10^{-8}$

$(0.941)^{95}\approx e^{95\times\ln(0.941)}\approx e^{95\times(- 0.061)}\approx e^{-5.795}\approx0.0031$

Then $P(X = 5)\approx75287520\times7.14\times10^{-8}\times0.0031\approx75287520\times2.2134\times10^{-10}\approx0.0166$ (but the hand - written note has a value around 0.17, which might be due to a wrong value of $p$ or miscalculations. If we use $p = 0.17$ (from the note) by mistake, the calculation would be different. However, following the correct binomial approach with $p\approx0.058$ - $0.059$, the probability is around 0.016 - 0.017 (or with the wrong $p$ as in the note, it can be around 0.17 due to miscalculating $p$ as 0.059 instead of a correct or incorrect $p$ value))

Part (a) Answer:

The probability that a randomly selected Amazon Prime member purchases at least ten items per month is approximately $\boldsymbol{0.059}$ (or more accurately around 0.058 - 0.059)

Part (b) Answer:

The random variable $T$ is defined as the number of Amazon Prime members in the sample of 100 who purchase at least ten items per month.

Part (c) Answer:

Yes, the random variable $T$ can be considered a binomial random variable. The reasons are:
1. There are a fixed number of trials ($n = 100$), where each trial is whether a particular Amazon Prime member purchases at least ten items per month.
2. Each trial has two outcomes: success (purchases at least ten items) or failure (purchases less than ten items).
3. The probability of success ($p$) is the same for each trial (the probability that a single member purchases at least ten items, which we calculated in part (a)).
4. The trials are independent (the purchasing behavior of one member does not affect the purchasing behavior of another member).

Part (d) Answer:

Using the binomial probability formula with $n = 100$, $p\approx0.059$ and $x = 5$, the probability is approximately $\boldsymbol{0.017}$ (or around 0.17 if there was a miscalculation of $p$ in the note)

Answer:

Step 1: Identify the distribution and parameters

We know that the number of items purchased per month for an individual member follows a normal distribution with mean $\mu = 6.4$ and standard deviation $\sigma=2.3$. We want to find $P(X\geq10)$, where $X$ is the number of items purchased per month. First, we calculate the z - score using the formula $z=\frac{x - \mu}{\sigma}$.

For $x = 10$, the z - score is $z=\frac{10 - 6.4}{2.3}=\frac{3.6}{2.3}\approx1.57$

Step 2: Find the probability using the z - table

We know that $P(X\geq10)=1 - P(X\lt10)$. Since $X$ is normally distributed, $P(X\lt10)=\Phi(z)$ (where $\Phi(z)$ is the cumulative distribution function of the standard normal distribution). Looking up $z = 1.57$ in the standard normal table, we find that $\Phi(1.57)\approx0.9418$.

So $P(X\geq10)=1 - 0.9418 = 0.0582\approx0.058$ (or approximately 0.059 as in the hand - written note, due to possible differences in z - table precision or rounding during z - score calculation)

For part (d), if we assume that the number of members who purchase at least 10 items follows a binomial distribution with $n = 100$ (number of trials), $p=P(X\geq10)\approx0.058$ (or 0.059 as in the note) and $x = 5$ (number of successes). The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$

$C(100,5)=\frac{100!}{5!(100 - 5)!}=\frac{100\times99\times98\times97\times96}{5\times4\times3\times2\times1}=75287520$

$p = 0.059$, $1-p = 0.941$, $k = 5$, $n = 100$

$P(X = 5)=75287520\times(0.059)^{5}\times(0.941)^{95}$

First, $(0.059)^{5}\approx0.059\times0.059\times0.059\times0.059\times0.059\approx0.059^{5}\approx7.14\times10^{-8}$

$(0.941)^{95}\approx e^{95\times\ln(0.941)}\approx e^{95\times(- 0.061)}\approx e^{-5.795}\approx0.0031$

Then $P(X = 5)\approx75287520\times7.14\times10^{-8}\times0.0031\approx75287520\times2.2134\times10^{-10}\approx0.0166$ (but the hand - written note has a value around 0.17, which might be due to a wrong value of $p$ or miscalculations. If we use $p = 0.17$ (from the note) by mistake, the calculation would be different. However, following the correct binomial approach with $p\approx0.058$ - $0.059$, the probability is around 0.016 - 0.017 (or with the wrong $p$ as in the note, it can be around 0.17 due to miscalculating $p$ as 0.059 instead of a correct or incorrect $p$ value))

Part (a) Answer:

The probability that a randomly selected Amazon Prime member purchases at least ten items per month is approximately $\boldsymbol{0.059}$ (or more accurately around 0.058 - 0.059)

Part (b) Answer:

The random variable $T$ is defined as the number of Amazon Prime members in the sample of 100 who purchase at least ten items per month.

Part (c) Answer:

Yes, the random variable $T$ can be considered a binomial random variable. The reasons are:

1. There are a fixed number of trials ($n = 100$), where each trial is whether a particular Amazon Prime member purchases at least ten items per month.
2. Each trial has two outcomes: success (purchases at least ten items) or failure (purchases less than ten items).
3. The probability of success ($p$) is the same for each trial (the probability that a single member purchases at least ten items, which we calculated in part (a)).
4. The trials are independent (the purchasing behavior of one member does not affect the purchasing behavior of another member).

Part (d) Answer:

Using the binomial probability formula with $n = 100$, $p\approx0.059$ and $x = 5$, the probability is approximately $\boldsymbol{0.017}$ (or around 0.17 if there was a miscalculation of $p$ in the note)