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:
At her current rate it will take here 70 minutes to finish the race
Simply it as follows:
1 + 39 = 40
2 + 38 = 40 etc
18 + 22 = 40
19 + 21 = 40
There are 19 of these plus 40 itself . . .therefore 20 of them
20 x 40 = 800
And finally the last number not used, 20
Therefore 800 + 20 = 820
Well you do 57 time 7 and get 399 and then it says how far will she have driven so you would add the 60 and get 459 miles.
