Question

According to a survey, only 15% of customers who visited the order page of a major online store made a purchase. Random samples of size 50 are selected. The mean of all the sample proportions of customers who will make a purchase after visiting the order page is? (Round this answer to 2 decimals.)

The standard deviation of all the sample proportions of customers who will make a purchase after visiting the order page is?

What proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the order page?

90% of the samples will have less than what percentage of customers who will make a purchase after visiting the order page? (Round this answer to 2 decimals.)

90% of the samples will have more than what percentage of customers who will make a purchase after visiting the order page? (Round this answer to 2 decimals.)

Answer 1

Sampling Distribution of the sample proportion is :

$\dpi{100} Mean = p$

$\dpi{100} Standard\ deviation(SD) = \sqrt{\frac{p(1-p)}{n}}$

The mean of all the sample proportions of customers who will make a purchase after visiting the order page = p = 15% = 0.15

The standard deviation of all the sample proportions of customers who will make a purchase after visiting the order page =

$\dpi{100} \small = \sqrt{\frac{0.15(1-0.15)}{50}} = 0.0505$

Proportion of the samples will have between 20% and 30% of customers who will make a purchase after visiting the order page will be :

$\small P(20 \% <p<30 \% ) = P(0.20<p<0.30) = P\left ( \frac{0.20-mean}{SD}<\frac{p-mean}{SD}<\frac{0.30-mean}{SD} \right )$

$\small = P\left ( \frac{0.20-0.15}{0.0505}<Z<\frac{0.30-0.15}{0.0505}\right )$

$\small = P\left ( 0.99<Z<2.97\right ) = P(Z<2.97) - P(Z<0.99)$

Using Standard normal tables or using the Excel function " NORMSDIST() "

$\small =0.9985 - 0.83895 = 0.1596$

Let 90% of the samples will have less than P₉₀ percentage of customers who will make a purchase after visiting the order page

So, we need to find P₉₀ such that :

$\small P(p<P_{90}) = 0.90$

$\small P\left ( \frac{p-mean}{SD}< \frac{P_{90}-mean}{SD} \right ) = 0.90$

$\small P\left ( Z< \frac{P_{90}-0.15}{0.0505} \right ) = 0.90$

Using Standard normal tables we get :

$\small \frac{P_{90}-0.15}{0.0505} = 1.28155$

$\small P_{90} = 1.28155*0.0505&plus;0.15 = 0.2147 = 0.21\ \ (2\ decimal\ place)$

90% of the samples will have less than 21.47 % of customers who will make a purchase after visiting the order page.

Let 90% of the samples will have more than P₁₀ percentage of customers who will make a purchase after visiting the order page

So, we need to find P₁₀ such that :

$\small P(p>P_{10}) = 0.90$

$\small P\left ( \frac{p-mean}{SD}> \frac{P_{10}-mean}{SD} \right ) = 0.90$

$\small P\left ( Z> \frac{P_{10}-0.15}{0.0505} \right ) = 0.90$

$\small P\left ( Z< \frac{P_{10}-0.15}{0.0505} \right ) = 0.10$

Using Standard normal tables we get :

$\small \frac{P_{10}-0.15}{0.0505} = -1.28155$

$\small P_{10} = -1.28155*0.0505&plus;0.15 = 0.0853 = 0.09\ \ (2\ decimal\ place)$

90% of the samples will have more than 8.53 % of customers who will make a purchase after visiting the order page.