6. 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 unusual to get 568 consumers who recognize the Dull Computer Company name? Find the mean and standard deviation to answer this question. Must show work and explain why or why not.

Given Information:

Population = n = 800

Probability = p = 68% = 0.68

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.

Step-by-step explanation:

Let us first find 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

we know that 68% of data fall within 2 standard deviations from the mean

μ ± 2σ = 544-2*13.2, 544+2*13.2

μ ± 2σ = 544-26.4 , 544+26.4

μ ± 2σ =  517.6, 570.4

μ ± 2σ = 518, 570

We can say with 68% confidence that 568 lies in the range of (518, 570) therefore, it would not be unusual to get 568 consumers who recognize the Dull Computer Company name.