The First National Bank of Wilson has 600 checking account customers. A recent sample of 50 of these customers showed 29 have a visa card with the bank. Construct the 90% confidence interval for the proportion of checking account customers who have a visa card with the bank.

0.58 +/- 0.115

(0.465, 0.695)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 29/50 = 0.58

Number of samples n = 50

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.58 +/- 1.645√(0.58(1-0.58)/50)

0.58 +/- 1.645(0.069799713466)

0.58 +/- 0.114820528652

0.58 +/- 0.115

(0.465, 0.695)

The 90% confidence interval estimate of the true population proportion of account customers who have a visa card with the bank is;

0.58 +/- 0.115

(0.465, 0.695)

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.