Answer:
the probability that a sample of the 35 exams will have a mean score of 518 or more is <em> 0.934 </em>or<em> 93.4%</em>.
Step-by-step explanation:
This is s z-test because we have been given a sample that is large (greater than 30) and also a standard deviation. The z-test compares sample results and normal distributions. Therefore, the z-statistic is:
(520 - 518) / (180/√35)
= 0.0657
Therefore, the probability is:
P(X ≥ 0.0657) = 1 - P(X < 0.0657)
where
- X is the value to be standardised
Thus,
P(X ≥ 0.0657) = 1 - (520 - 518) / (180/√35)
= 1 - 0.0657
= 0.934
Therefore, the probability that a sample of the 35 exams will have a mean score of 518 or more is <em>0.934 or 93.4%</em>.
Answer:
Step-by-step explanation:
Given that a teacher gives a test to a large group of students. The results are closely approximated by a normal curve
mu =74 and sigma =8
A grade starts from 100-8 = 92nd percentile
Z score for 92nd percentile = 1.405
X score = 74+8(1.405) = 85.24
--------------------
B cut off is to next 16%
Hence C would start for scores below 100-(8+16) = 76%
76th percentile = 0.705*8+74 =79.64
Total servings to cups is 7:4
means total servings is 7 and total cups is 4
total=cups+cones
therefor
cones=total-cups
we want servings to cones
we know that cups=4 units and total =7
we apply
cones=total-cups
cones=7-4
cones=3
cones=3 units
therefor, serrvings to cones is 7:3
answer is C
Answer:
13,8%
Step-by-step explanation:
There are six employees and six cheks, so there are 36 (6x6) possible combinations so if we need to measure the probability that five of them receive the exact check is only one for each one of them over the 36 possibilities, so 1/36 for one plus 1/36 the second and so on. 5/36 = 13,8%.
