A survey of 225 students showed the mean number of hours spent studying per week was 20.6 and the standard deviations was 2.7
2 answers:
Answer:
The margin of error is approximately 0.3
Step-by-step explanation:
The following information has been provided;
The sample size, n =225 students
The sample mean number of hours spent studying per week = 20.6
The standard deviation = 2.7
The question requires us to determine the margin of error that would be associated with a 90% confidence level. In constructing confidence intervals of the population mean, the margin of error is defined as;
The product of the associated z-score and the standard error of the sample mean. The standard error of the sample mean is calculated as;
where sigma is the standard deviation and n the sample size. The z-score associated with a 90% confidence level, from the given table, is 1.645.
The margin of error is thus;
Therefore, the margin of error is approximately 0.3
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The correct answer is A.
Answer:
The chance of getting exactly 3 hits is = 0.20
Step-by-step explanation:
P.S - The exact question is -
As given,
F(x) = 0 , x < 1
0.30 , 1 ≤ x < 2
0.56 , 2 ≤ x < 3
0.76 , 3 ≤ x < 4
0.9 , 4 ≤ x < 5
1 , 5 ≤ x
Now,
f(x) = 0.30 , x = 1
0.56 - 0.30 = 0.26 , x = 2
0.76 - 0.56 = 0.20 , x=3
0.9 - 0.76 = 0.14 , x = 4
1 - 0.9 = 0.1 , x = 5
0, otherwise
Now,
The chance of getting exactly 3 hits is = f(x = 3) = 0.20
