Twelve of the 20 students in Mr. Skinner’s class brought lunch from home. Fourteen of the 21 students in Ms. Cho’s class brought
2 answers:
Solution:
Number of students in Mr.Skinner's class who brought lunch from home if there are 20 students in the class=12
Fraction of students who brought lunch from home in Mr. Skinner's class=
Number of students in Ms. Cho's class who brought lunch from home if there are 21 students in the class=14
Fraction of students who brought lunch from home in Ms. Cho's class=
As Siloni is using two 15-section spinners to simulate randomly selecting students from each class and predicting whether they brought lunch from home or will buy lunch in the cafeteria.
Number of Congruent sectors in each Spinner=15
So, if we represent students from Mr. Skinner's class who brought lunch from home in Spinner having 15 congruent Sectors =
So, if we represent students from Mrs. Cho's class who brought lunch from home in Spinner having 15 congruent Sectors =
Mr Skinner class +1 = Mr's Cho's Class
So Ms Cho's class =One more sector of the Skinner-class spinner will represent bringing lunch from home.
Option A which is One more sector of the Skinner-class spinner will represent bringing lunch from home represents Ms Cho's Class.
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Answer:
The confidence interval for the mean is given by the following formula:
(1)
The 90% confidence interval for this case would be (38.01, 44.29) and is given.
The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .
And the best option would be:
The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29
Step-by-step explanation:
Assuming this complete question: Which statement gives a valid interpretation of the interval?
The store manager is 90% confident that the average amount spent by the 36 sampled customers is between S38.01 and $44.29.
There is a 90% chance that the mean amount spent by all customers is between S38.01 and $44.29.
There is a 90% chance that a randomly selected customer will spend between S38.01 and $44.29.
The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
The confidence interval for the mean is given by the following formula:
(1)
The 90% confidence interval for this case would be (38.01, 44.29) and is given.
The best interpretation for this case would be: We are 90% confident that the true average is between $ 38.01 and $ 44.29 .
And the best option would be:
The store manager is 90% confident that the average amount spent by all customers is between S38.01 and $44.29
The Given Sequence is an Arithmetic Sequence with First term = -19
⇒ a = -19
Second term is -13
We know that Common difference is Difference of second term and first term.
⇒ Common Difference (d) = -13 + 19 = 6
We know that Sum of n terms is given by :
Given n = 63 and we found a = -19 and d = 6
The Sum of First 63 terms is 10521