Ms. Ringer asked her class, "What are the favorite colors of the students in our school?" Which student correctly describes Ms.
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We are given the following data:
Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22
We are to find the confidence intervals for 90%, 95% and 98% confidence level.
Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.
Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 90% confidence interval will be (110.69, 136.51)
Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 95% confidence interval will be (108.22, 138.98)
Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 98% confidence interval will be (105.34, 141.86)
Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)
As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22
We are to find the confidence intervals for 90%, 95% and 98% confidence level.
Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.
Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 90% confidence interval will be (110.69, 136.51)
Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 95% confidence interval will be (108.22, 138.98)
Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327
Lower end of confidence interval =
Using the values, we get:
Lower end of confidence interval=
Upper end of confidence interval =
Using the values, we get:
Upper end of confidence interval=
Thus the 98% confidence interval will be (105.34, 141.86)
Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)
As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
Answer:
The correct answer is option b)
50a+10b<1000; 200a+360b < 7200; a> 0; b>0
Step-by-step explanation:
We are given that Two kind of crated cargo namely A and B to be shipped by truck.
<u>Cargo A:</u>
Volume of each crate of cargo A = 50 cubic ft
Weight of each crate of cargo A = 200 pounds
Let number of crates of cargo A to be shipped = a
Total volume of 'a' crates of cargo A = 50a cubic ft
Total weight of 'a' crates of cargo A = 200a pounds
<u>Cargo B:</u>
Volume of each crate of cargo B = 10 cubic ft
Weight of each crate of cargo B = 360 pounds
Let number of crates of cargo B to be shipped = b
Total volume of 'b' crates of cargo B = 10b cubic ft
Total weight of 'b' crates of cargo B = 360b
Total volume allowed in the truck is 1000 cubic ft
Total volume of 'a' crates of Cargo A and Total volume of 'b' crates of Cargo B = 50a+10b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)
So, the inequality becomes
....... (1)
Total weight allowed (load limit) in the truck is 7200 pounds
Total weight of 'a' crates of Cargo A and Total weight of 'b' crates of Cargo B = 200a+360b cubic ft (This sum should be less than volume of truck so that it can fit in the truck)
So, the inequality becomes
....... (1)
And number of crates of cargo A and B are always a positive number.
So, a > 0 and b > 0.
So, the correct answer is option b.
<em>b. 50a+10b<1000; 200a+360b < 7200; a> 0; b>0</em>