Complete question :
Suppose there are n independent trials of an experiment with k > 3 mutually exclusive outcomes, where Pi represents the probability of observing the ith outcome. What would be the formula of an expected count in this situation?
Answer: Ei = nPi
Step-by-step explanation:
Since Pi represents the probability of observing the ith outcome
The number of independent trials n = k>3 :
Expected outcome of each count will be the product of probability of the ith outcome and the number of the corresponding trial.
Hence, Expected count (Ei) = probability of ith count * n
Ei = nPi
-4j -1 -4j + 6 can be arranged by the commutative property of addition.
-4j - 4j -1 + 6
If we combine like terms we get -8j +5
Answer: A:60%
Step-by-step explanation:
Since the scores for a golf tournament are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = scores for the tournament.
µ = mean score
σ = standard deviation
From the information given,
µ = 210
σ = 80
We want to find the probability percent of golfers that scored less than Ella. It is expressed as
P(x < 230)
z = (230 - 210)/80 = 0.25
Looking at the normal distribution table, the probability corresponding to the z score is 0.5987
Therefore, the percent of golfers that scored less than Ella is
0.5987 × 100 = 59.87
Approximately 60%
Answer:
I got 70.56
Step-by-step explanation:
