Answer:
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
That is z with a pvalue of 
, so Z = 1.645.
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 34.09 - 3.25 = 30.84.
The upper end of the interval is the sample mean added to M. So it is 34.09 + 3.25 = 37.34.
A 90% confidence interval for the mean number of letter sounds identified in one minute is: (30.84, 37.34).
To solve this, we use the z test.
The formula:
z = (x – u) / s
where x is sample value = 20, u is the mean = 15, and s is
the standard deviation = 2.5
z = (20 – 15) / 2.5
z = 2
Since we are looking for values greater than 20, this is
right tailed test. We use the standard distribution tables to find for P.
P = 0.0228
Therefore:
number of students = 100 * 0.0228 = 2.28
<span>2 to 3 students will get greater than 20 measurement</span>
36 throws because you divide 60 by five and get 12 and then multiply 3 by 12 giving you 36, are you sure this is high school
Answer: 2
explanation: they’re both one place down
I believe it is written as 0.43 <span />
