A survey showed that 76% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 8 adults are randomly selected, find the probability that at least 7 of them need correction for their eyesight. Is 7 a significantly high number of adults requiring eyesight correction?

as probability of this is higher than 0.05 level this is not significantly high number .

Step-by-step explanation:

we have here a binomial distribution with the following data:

p = 0.76 and n = 8

therefore probability that a least 7 of the 8 adults require eyesight correction is P (X> = 7)

 P (X> = 7) = P (X = 7) + P (X = 8) = nCr * (p) ^ (n) * (1-p) ^ (n-r)

= (8C7) * (0.76) ^ 8 * (0.24) ^ (8-7) + (8C8) * (0.76) ^ 8 * (0.24) ^ (8-8)

nCr = n! / (r! * (n-r)!

replacing:

(8C7) = 8! / (7! * (8-7)!) = 8

(8C8) = 8! / (8! * (8-8)!) = 1

P (X> = 7) = 8 * 0.0267 + 1 * 0.11130 = 0.325

as probability of this is higher than 0.05 level this is not significantly high number .

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft

54 it just is

step-by-step explanation: