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
= sqrt(p(1-p)/n)
= sqrt(090(1-090)/144)
= 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,
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:-
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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