Douglas wants to estimate the percentage of homeowners that own at least one dog. He surveys 125 randomly selected homeowners to
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The correct answer is A.) { - 0.79 ; 1.08 }, because
<span><span> <span>For </span><span>ax^2 + bx + c = 0</span><span>, the value of </span>x<span> is given by:</span></span> <span> ;
a = - 7 ; b = 2 ; c = 6 ;
b^2 - 4ac = 4 + 168 = 172 ; </span></span>
≈ 13.11 ;<span><span>
x 1 = ( - 2 + 13.11 ) / (- 14) = 11.11 / ( - 14 ) </span></span>≈<span><span> -0.79 ;
x = 2 = ( - 2 - 13.11 ) / ( - 14 ) = (-15.11) / (- 14 ) </span></span>≈ 1.08 ;<span><span>
</span></span>
<span><span> <span>For </span><span>ax^2 + bx + c = 0</span><span>, the value of </span>x<span> is given by:</span></span> <span> ;
a = - 7 ; b = 2 ; c = 6 ;
b^2 - 4ac = 4 + 168 = 172 ; </span></span>
x 1 = ( - 2 + 13.11 ) / (- 14) = 11.11 / ( - 14 ) </span></span>≈<span><span> -0.79 ;
x = 2 = ( - 2 - 13.11 ) / ( - 14 ) = (-15.11) / (- 14 ) </span></span>≈ 1.08 ;<span><span>
</span></span>
Answer:
b) 0.2961
c) 0.2251
d) Mean = 11.25, Standard deviation = 1.667
Step-by-step explanation:
We are given the following information:
We treat trucks undergoing a brake inspection passin as a success.
P( trucks undergoing a brake inspection passes the test) = 75% = 0.75
a) Conditions for binomial probability distribution
- There are n independent trial.
- Each trial have a success probability p
- The probability of success is same for all trials.
Then the number of trucks undergoing a brake inspection follows a binomial distribution, where
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15
b) P(proportion of groups will between 8 and 10 trucks pass the inspection)
We have to evaluate:
c) P( exactly 3 trucks fail the inspection)
p = 0.25
d) Mean and standard deviation