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)

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.