Answer:
1, 2, 3, 6, 103, 206, 309, 618
Answer is From all those aforementioned, the true statement is: Thanh’s grade is 2.5 standard deviations below the mean of the test grades.
explanation
The z scores are standard deviations. If, for example, a tool returns a z score of +2.5, it means that the result is a standard deviation of 2.5. Very high or very low (negative) z scores associated
with very low p values are at the ends of the normal distribution.
The confidence interval formula is computed by:
Xbar ± Z s/ sqrt (n)
Where:
Xbar is the mean
Z is the z value
S is the standard deviation
N is the number of samples
So our given are:
90% confidence interval with a z value of 1.645
Sample size 40, 45
Mean 180, 179
Standard deviation 2, 4
So plugging that information in the data will give us a
confidence interval:
For 1:
Xbar ± Z s/ sqrt (n)
= 180 ± 1.645 (2 / sqrt (40))
= 180 ± 1.645 (0.316227766)
= 180 ± 0.520194675
= 179.48, 180.52
For 2:
Xbar ± Z s/ sqrt (n)
= 179 ± 1.645 (4 / sqrt (45))
<span>= 179 ± 1.645 (0.596284794)</span>
therefore, the answer is letter b
Answer:
66 ≤ f ≤100
Explanation
Mean= ( Σ x ) / n
Mean= sum of scores/ number of subject she took
Now, she already too 3 subject which sum is 85+83+86=254
Now we need to know range of score for her to have (grade) a mark between 80 and 89
Now let take the lower limit mean=80
The lowest score she can get is
Mean = ( Σx) / n
80=(85+83+86+f)/4
80×4= 254+f
Therefore, f= 320-254=66
Therefore the minimum score she can have to have a B is 66.
Then, let take the upper limit mean 89. i.e the maximum she can have so that she don't have an A grade.
Mean = ( Σx) / n
89=( 83+85+86+f)/4
89×4= 254+f
f= 356-254
f=102.
Therefore this shows that she cannot have an A grade in the exam. The maximum score for the exam is 100.
There the range of score is 66 ≤ f ≤100 to have a B grade
66 ≤ f ≤100 answer
Since she cannot score 102 in the examination.
Answer:
1/9
Step-by-step explanation:
For these questions, you have to first find the probability of each event individually occuring then multiply these two probabilities together. So, lets first find the probability of Event 1 occuring (rolling a number higher than 1). On a dice, we know that there are 6 numbers (1, 2, 3, 4, 5, 6) of which the numbers 5 and 6 are higher than 4. Therefore, the proability of Event 1 occuring is 2/6 = 1/3. Now, we find the probability of event 2 occuring.
We know that there are 3 options on the spinner that the arrow can land on. It is also evident that they all have the same chance of landing on it. Therefore, the possibility is 1/3.
Now that we have both of our probabilities, we multiply the answer together.
Therefore, our answer will be:
1/3 x 1/3, which gives 1/9.
