Based on a nationwide poll, a seller of printed advertisements estimates
that 56% of all adults usually open all the mail they receive. If this is still the current rate at which adults open mail, what is the probability that in a random sample of 1000 adults, the number who usually open all of their mail will be
(a) Less than 541
(b) 570 or more
Answers
Given : Based on a nationwide poll, a seller of printed advertisements estimates that 56% of all adults usually open all the mail they receive.
a random sample of 1000 adults,
To Find :
probability of numbers of people who usually open all of their mail will be
(a) Less than 541
(b) 570 or more
Solution:
56% of all adults usually open all the mail they receive
=> p = 0.56
q = 1 - p = 1 - 0.56 = 0.44
n = 1000
Mean = np = 1000 x 0.56 = 560
Variance = npq
= 1000 * 0.56 * 0.44
= 246.4
SD = √Variance = √ 246.4 = 15.7
Z score = ( Value - Mean)/SD
Less than 541
Z score = ( 541 - 560)/15.7 = -1.21 => probability = 0.113
1000 x 0.113
Hence 113 People are less than 541
570 or more
Z score = ( 570 - 560)/15.7 = 0.637 = proportion 0.738
1 - 0.738 = 0.262 probability
0.262 x 1000 = 262
Hence 262 people 570 or more
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