Aband is interested in determining the mean credit card balance for its customers. from a sample of 70 customers, the bank determine: x bar = $9312, sample standard deviation = $4007 calculate the 95% confidence interval.

($8349.69, $10274.31)

Step-by-step explanation:

When population standard deviation is unknown, then the formula to find the confidence interval for population mean is given by :-

, where  n= sample size

s= sample standard deviation.

= sample mean

As per given , we have

Significance level :

Degree of freedom : df = 69

Using t-distribution ,

Critical value for confidence interval :

s= $4007

Now, required confidence interval for the mean credit card balance for its customers will be :-

The 95% confidence interval for the mean credit card balance for its customers= ($8349.69, $10274.31)

Step-by-step explanation:

step explanation:

Assuming the credit card balance follows normal distribution,

We would determine a 95% confidence interval for the mean credit card balance.

Number of samples. n = 70

Mean, u = 9312

Standard deviation, s = 4007

For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

9312 ± 1.96 × 4007/√70

= 9312 ± 1.96 × 478.93

= 9312 ± 938.76

The lower end of the confidence interval is 9312 - 938.76 = 8373.24

The upper end of the confidence interval is 9312 - 938.76 =10250.76

Therefore, with 95% confidence interval, the mean credit card balance is between $8373.24 and $10250.76

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

you can't solve x because it is 2 linear equation and you only have one equation