B. the middle half of the data set
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:
c
Step-by-step explanation: If you do the math you can clearly see and I got the answer correct.
Answer:
Step-by-step explanation:
y = -3x - 3
m = -3_______
b = -3________
y = 2x + 2
m = 2_______
b = _2_______
SOLVE NOW : -3x - 3 = 2x + 2
- 5x = 5
x = -1
subst in y : y = 2(-1)+2 = 0
the solution is : ( - 1 , 0 )
y = -3x- 3 an equation for the line 'red'
y = 2x+2 an equation for the line 'bleus'
