Respuesta :
Answer:
672.7436 units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale
Step-by-step explanation:
No. of customers pass the display on the day of sale = 1462
Proportion of people will purchase if pass such a special display = 0.43
Mean = E(X)=np
E(X)=[tex]1462 \times 0.43[/tex]
E(X)=628.66
Standard deviation = [tex]\sqrt{npq}[/tex]
Standard deviation =[tex]\sqrt{1462 \times 0.43 \times (1-0.43)}=18.92[/tex]
Now we are supposed to find how many units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale
P(X<x)=0.01
Using z table
[tex]P(\frac{x-\mu}{\sigma})= 2.33 \\\frac{x-628.66}{18.92}=2.33\\x=(2.33 \times 18.92)+628.66\\x=672.7436[/tex]
Hence 672.7436 units of the sale item should the store stock in order to have at most a 1 percent chance of running short of the item on the day of the sale
The number of units should be 672.7436 units.
Calculation of number of units:
Since No. of customers pass the display on the day of sale is 1462 and the proportion of people who will purchase if pass such a special display is 0.43
So,
Mean should be = 43% of 1,462
= 628.66
Now the standard deviation should be
[tex]= \sqrt{1462 \times 0.43 \times (1-0.43)}[/tex]
= 18.92
Now the number of units should be
[tex]x - 628.66 \div 18.92 = 2.33[/tex]
x = 672.7436
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