Question

Problem 7: 13 points) In a restaurant, the proportion of people who order coffee with their dinner is 0.9.A sample of 144 clients of the restaurant is taken. a. What are the expected value, standard deviation and the shape of the sampling distribution of p (the sample proportion). 1s4LE b. What is the probability that the sample proportion of people who will order coffee with their meal is between 0.85 and 0,875 2 c. What is the probability that the sample proportion of people who will order coffee with their meal is at least 0.945 11시

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