The probability that of those 45 people sampled, between 33 and 36 of them own a cell phone is 0.2413
How to solve for the probability
We have to solve for the probability that of those 45 people sampled, between 33 and 36 of them own a cell phone
The standard deviation is given as
[tex]s = \sqrt{} \frac{P(1-p)}{n}[/tex]
where we have p = proportion = 70% = 0.7
q = 1 - p
= 1 - 0.7
= 30 % = 0.3
s = [tex]\sqrt{\frac{0.7(1-0.7}{45} }[/tex]
= 0.07
we are to find the interval of
P(33 ≤ x ≤ 36)
this would be written as
(33 / 45 - 0.70) / 0.07 < z < (36 / 45 - 0.70) / 0.07)
p(0.49 < z < 1.46)
we have to find the critical values of p(z < 0.49) and p(z < 1.46)
= 0.9284 - 0.68721
= 0.2413
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