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 $102.5 is 0.0918.
No, because 11.6/2 is 5.8 so 5.8 is half of the tank. if it uses 1.45 per hour and it's used for 3.5 hours there will only be 5.075
Answer:
The price of one notebook is 2$
Step-by-step explanation:
3x + 2= 4x
Take x to the same side —> 4x-3x=2 —> x=2
Answer:
Interpret and compare the data.
Step-by-step explanation:
elim options 1 and 2
analyzing means you don't have findings
Answer:
um i can't really understand what your trying to do here
Step-by-step explanation:
