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Problem 7: 13 points) In a restaurant, the proportion of people who order coffee with their dinner is 0.9.A sample of 144 cli

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Answer #1

Answer:

Given that

p = 0.90

samlpe n = 144

a)

To determine the expected value, standard deviation and shape of distribution of p

we know,

E(x) = n × p

substitute values in above expression

E(x) = 144 × 0.90

E(x) = 129.6

Standard error can be given as follows

\sigma = sqrt(p(1-p)/n)

\sigma = sqrt(090(1-090)/144)

\sigma = 0.025

Shape is approximately normal

(b)

To determine probability of mean between the 0.85 and 0.875

p1 = 0.85

p2 = 0.875

By applying normal distribution:-

Consider,

P-p p(1-p)

z(p1 = 0.85) = - 2

z(p2 = 0.875) = - 1

So

P( - 2 < z < -1) = P(z > -2) - P(z > -1)

P( - 2 < z < -1) = 0.9772 - 0.8430 [Since from the standard normal distribution]

P( - 2 < z < -1) = 0.1342

(c)

To determine the probability that the proportion of people who will order with their meal is atleast 0.945

Consider,

p = 0.945

By applying normal distribution:-

P-p p(1-p)

By substituting the known values in the above expression we get Z value

i.e.,

z = 1.8

P(z > 1.8) = 0.0359 [Since from the normal distribution table]

Hence

P(z > 1.8) = 0.0359

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