Usually about 65% of the patrons of a restaurant order burgers. A restaurateur a

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Question

Usually about 65% of the patrons of a restaurant order burgers. A restaurateur anticipates serving about 165 people on Friday. Let X be the numbers of burgers ordered on Friday. Then X is binomially distributed with parameters n = 165 and p = 0.65.

What is the expected number of burgers (X) ordered on Friday?

Find the standard deviation of X (X)?

If the restaurant ordered meats to prepare about 115 burgers for Friday evening. Use normal approximation of binomial distribution to find the probability that on Friday evening some orders for burgers from the patron cannot be met.

How many burgers the restaurant should prepare beforehand so that the chance that an order of burger cannot be fulfilled is at most 0.04? i.e. Find a such that P(X > a) = 0.04 using normal approximation of binomial distribution.

Explanation / Answer

What is the expected number of burgers (X) ordered on Friday?

Expectation = np =165*0.65 = 107.25

Find the standard deviation of X (X)?

Variance = np(1 - p) = 37.5375

Standard deviation = 6.1268

Z value for 115, z =(115-107.25)/6.1268 = 1.26

P( x >115) = P( z >1.26)

=0.1038

From standard normal distribution, P( z >1.751) =0.04

X=mean+z*sd

=107.25+1.751*6.1268

=117.98

=118 ( rounded to whole number)

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