A stadium designer has placed 900 seats in one section of the stadium. Assume that the weights of people who will occupy those seats are i. i.d. with a mean of 160 pounds and an SD of 25 pounds. Fill in the blanks with numbers and show your calculation:
With chance about 99%, the total weight of people in the 900 seats will be in the range +- pounds.

Required confidence interval = (142,065, 145,935)

Step-by-step explanation:

The mean of 900 people is 160 pounds

And the standard deviation = 25 pounds.

We can find the confidence interval for the weights and then multiply by 900 to obtain the required confidence interval for all the possible total weights for the stadium.

Confidence Interval is basically an interval of range of values where the true mean can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Mean) ± (Margin of error)

Sample Mean = 160 pounds

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error of the mean)

Critical value will be obtained using the z-distribution. This is because this 900 people represent a sample size that is large enough for t-score = z-score

Critical level for 99% significance level = 2.58

Standard error of the mean = σₓ = (σ/√n)

σ = standard deviation of the sample = 25

n = sample size = 900

σₓ = (25/√900) = 0.8333

99% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]

CI = 160 ± (2.58 × 0.8333)

CI = 160 ± 2.15

99% CI = (157.85, 162.15)

99% Confidence interval = (157.85, 162.15)

For the total 900 seats, the confidence interval for total weights would be

900 × (157.85, 162.15) = (142,065, 145,935)

Hope this Helps!!!

ans789

step-by-step explanation:

the answer would be 0.54