Answer:
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Step-by-step explanation:
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:
point s at (0,5) point t at (5,3) point u at (4,1)
Step-by-step explanation:
happy life
Answer:
plus or minus square root 50
Step-by-step explanation:
x^2 = 50
Take square root of both sides
x = +/ - √50
Answer:
plus or minus square root 50
We are given with
initial consumption = 13.1
consumption after 17 years = 19.9
P < 0.01
The additional information are needed to compute for the confidence interval for the increase are
the standard deviation for the initial consumption and the increased consumption
