p = 0.68
n = 800
µ = np
µ = 800(.68) = 544
Standard deviation = √(np(1-p) )
= √(800(.68)(.32) )
= 13.1939
µ±2σ
(544 -2(13.193), 544+2(13.1939))
(517.6, 570.4)
554 is not unusual because it is in interval.
A survey for brand recognition is done and it is determined that 68% of consumers have heard of Dull Computer Company. If a survey of 800 randomly selected consumers is to be held would it be unusual for 554 of them to recognize Dull Computer Company?
Please show the work so I can understand what formula to use and work it out my self with the other problems.
1 answer
1 answer