Question

Based on 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 usually open all mail their mail will be . a . less than 541. b . 570 or more​

Answer 1

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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