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)

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt

table 3 represents a direct variation function. : )