A crowd of 28,760 fans will be attending a soccer game. On average 27% of attendees will want to buy a meat pie. Use a normal distribution to find how many meat pies the stadium should order to ensure a less than 20% chance of running out of meat pies. (Note that no σ has been provided. To determine it, we assume that we are approximating a binomial distribution in which σ 2 = np(1−p). Moreover, to correct for issues in approximating a discrete quanity (people) using a continuous distribution, it is normal to subtract 0.5 from lower bounds and add 0.5 to upper bounds.)

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Asked Oct 14, 2019
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A crowd of 28,760 fans will be attending a soccer game. On average 27% of attendees will want to buy a meat pie. Use a normal distribution to find how many meat pies the stadium should order to ensure a less than 20% chance of running out of meat pies. (Note that no σ has been provided. To determine it, we assume that we are approximating a binomial distribution in which σ 2 = np(1−p). Moreover, to correct for issues in approximating a discrete quanity (people) using a continuous distribution, it is normal to subtract 0.5 from lower bounds and add 0.5 to upper bounds.)

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Expert Answer

Step 1

Given,

population proportion of success is P = 27% = 0.27

n = Number of fans = 28760

Step 2

We have to find x where, P(X ≥ x) < 0.20

Here,

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The population mean is computed as: L=nxp 28760 x 0.27 = 7765.2 and the population standard deviation is computed as: Vnx px (1p) /28760 x 0.27 x (1 -0.27) o= 75.2901

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Step 3
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We get P(X2x) 0.20 P x 7765.2 75.2901 X- < 0.20 x-7765.2 P Z2 <0.20 75.2901

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