Solar-heat installations successfully reduce the utility bill 60% of the time, then the probability is P=0.04635.
We are aware that 60% of the time, solar heating installations successfully lower utility costs. We estimate the likelihood that at least 9 out of 10 solar-heat installations will be successful and result in lower utility costs.
So,
We can write,
p = 60%
p = 0.6
n=10.
We have a binomial distribution,
X : (10, 0.6).
We use the formula to calculate the probability:
P ( X = [tex]x[/tex] ) = [tex]\left[\begin{array}{c}n\\x\\\end{array}\right][/tex] * [tex]p^{x}[/tex] * [tex](1-p)^{n-x}[/tex]
Then,
P ( X ≥ 9 ) = 1 - P ( X < 9 )
P ( X ≥ 9 ) = 1 - P ( X < 8 )
P ( X ≥ 9 ) = 1 - Σx=0,8 P ( X = [tex]x[/tex] )
P ( X ≥ 9 ) = 1 - Σx=0,8 [tex]\left[\begin{array}{ccc}10\\x\\\end{array}\right][/tex] * [tex](0.6)^{x}[/tex] * [tex](1-0.6)^{10-x}[/tex]
P ( X ≥ 9 ) = 1 - ( 0.00011 + 0.00157 + 0.01062 + 0.04247 + 0.11148 + 0.20066 + 0.25082 + 0.21499 + 0.12093 )
P ( X ≥ 9 ) = 1 - 0.95365
P ( X ≥ 9 ) = 0.04635
Therefore,
Solar-heat installations successfully reduce the utility bill 60% of the time, then the probability is P=0.04635.
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