Answer:
1 quantitative
2 32 responded yes, 18 responded no
3 margin error 32%{n}
4 error of students on their phone / not
5 18/-/50
6 16%
7 2%
Step-by-step explanation:
a 95% confidence interval with a 4 percent margin of error means that your statistic will be within 4 percentage points of the real population value 95% of the time.
The margin of error tells you the range of values above and below a confidence interval.
Answer: 0.6827
Step-by-step explanation:
Given : Mean IQ score : 
Standard deviation : 
We assume that adults have IQ scores that are normally distributed .
Let x be the random variable that represents the IQ score of adults .
z-score : 
For x= 90

For x= 120

By using the standard normal distribution table , we have
The p-value : 

Hence, the probability that a randomly selected adult has an IQ between 90 and 120 =0.6827
Answer: An article was sold at a 20% discount, at Rs. 400. What was its marked price?
Let M denotes the required marked price of the given article.
Hence from above data we get following relation,
(1 - 20/100)*M = 400
or (4/5)*M = 400
or M = (5/4)*400 = 500 (Rs) [Ans]
Step-by-step explanation:
Answer:
ok ,so it. will be 4 ..................
Answer:
(a) 0.007238 or 07238%
(b) 0.003468 or 0.3468%
Step-by-step explanation:
(a) Since all it takes is one defective rivet for a seam to be reworked. The probability of a defective rivet 'p' for 16% of seams needing reworking is:
![1-(1-p)^{24} = 0.16\\1-p = \sqrt[24]{0.84}\\p=0.007238](https://tex.z-dn.net/?f=1-%281-p%29%5E%7B24%7D%20%3D%200.16%5C%5C1-p%20%3D%20%5Csqrt%5B24%5D%7B0.84%7D%5C%5Cp%3D0.007238)
The probability that a rivet is defective 0.007238 or 0.7238%.
(b) To ensure that only 8% of seams need reworking, the probability 'p' must be:
![1-(1-p)^{24} = 0.08\\1-p = \sqrt[24]{0.92}\\p=0.003468](https://tex.z-dn.net/?f=1-%281-p%29%5E%7B24%7D%20%3D%200.08%5C%5C1-p%20%3D%20%5Csqrt%5B24%5D%7B0.92%7D%5C%5Cp%3D0.003468)
In order to ensure that only 8% of all seams need reworking, the probability of a defective rivet should be 0.003468 or 0.3468%.