A sample of 100 consumers shows that the average amount spent annually on a debit card is $8000. And assume that the population standard deviation of debit card expenditure is $500. Construct a 95% confidence interval of the population mean amount spent annually on a debit card.

The confidence interval is ($7902, $8098)

Step-by-step explanation:

The formula for confidence interval of a normal distribution is given as:

Confidence Interval = μ ± z × σ/√n

Where:

μ is the mean amount spent annually = $8,000

σ is the standard deviation = 500

n is the number of samples = 100 customers

z = 95% confidence interval = 1.96

Confidence Interval = $8000 ± 1.96 ×500√100

= $8000 ± 1.96 × 500/10

Confidence Interval = $8000 ± 1.96 × 50

= $8000 ± 98

= $8000 - 98 = $7902

= $8,000 + 98 = $8098

Therefore, the 95% confidence interval of the population mean amount spent annually on a debit card = ($7902, $8098)

answer: the first graph

step-by-step explanation:

sorry

step-by-step explanation:

sorry, but im pretty sure its a now that i look at it