Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
Answer:
85.
Step-by-step explanation:
Answer: Company b
Step-by-step explanation:
because company B is only 5percent it is better and can pay in 8 years
Answer:
Mean = 35
Variance = 291.7
Step-by-step explanation:
Data provided in the question:
X : 1, 2, 3, 4, 5, 6
All the data are independent
Thus,
The mean for this case will be given as:
Mean, E[X] = 
or
E[X] = 
or
E[X] = 3.5
For 10 days, Mean = 3.5 × 10 = 35
And,
variance = E[X²] - ( E[X] )²
Now, for this case of independent value,
E[X²] = 
or
E[X²] = 
or
E[X²] = 
or
E[X²] = 15.167
Therefore,
variance = E[X²] - ( E[X] )²
or
variance = 15.167 - 3.5²
or
Variance = 2.917
For 10 days = Variance × Days²
= 2.917 × 10²
= 291.7
Answer:
55
Step-by-step explanation:
CE is a bisector so DE has the same measure as BE.
