A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. A survey of 800 randomly selected consumers is to be conducted. For such groups of 800, would it be significant to get 634 consumers who recognize the Dull Computer Company name? Consider as significant any result

Given Information:

Probability = p = 68% = 0.68

Population = n = 800

it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.

Step-by-step explanation:

We can check whether it would be significant to get 635 consumers who recognize the Dull Computer Company by finding out the mean and standard deviation.

mean = μ = np

μ = 800*0.68

μ = 544

standard deviation = σ = √np(1-p)

σ = √800*0.68(1-0.68)

σ = 13.2 ≈ 13

we know that 99% of data fall within 3 standard deviations from the mean

μ ± 3σ = 544+3*13, 544-2*13

μ ± 3σ = 544+39, 544-39

μ ± 3σ = 583, 505

So we can say with 99% confidence that the number of consumers who can  recognize the name of Dull Computer Company will be from 505 to 583 and since 583 < 634 we can conclude that it would not be significant to get 634 consumers who might recognize the name of Dull Computer Company.

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

source