Using the z-distribution, it is found that:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
<h3>What is the z-distribution confidence interval?</h3>
The confidence interval is:

In which:
is the difference between the population means.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of
, so the critical value is z = 1.96.
The estimate and the standard error are given by:

Hence the bounds of the interval are given by:


1.74 is outside the interval, hence:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
More can be learned about the z-distribution at brainly.com/question/25890103
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(x-10)(x-4) because 4-10=6 then multiply that by 0 from 4-4
Answer:
f=11
Mode=4
Median=4
Explanation:
We are given that
a.Mean of the exam score,
=3.5
Score(x) frequency C.F
1 1 1
2 3 4
3 f 15(4+f=4+11)
4 13 28
5 4 32



Using the formula





b.Mode:The number which is repeat most times .
4 repeat most times
Hence, mode of all exam scores=4
N=32
N is even

Median=
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