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:
All of the above
Step-by-step explanation:
10, 100, 1000, 100000 all are rational numbers because you can write them as:
10/1 , 100/1, 1000/1, 100000/1
Answer:
1) 102/21
2) 21/51
3) 21/24
Step-by-step explanation:
Answer:
Step-by-step explanation:
let me know if this is right please :(
Answer:
172 ft 2
Step-by-step explanation:
SA=2lw+2lh+2hw, for
A=2(lw + lh + hw)=2·(7 · 2+ 8 · 2 + 8 · 7)= 172 ft 2.
Hope this helps,
Best of Luck!
