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.
S is slope
S= rise/run
S1=3/12=1/4
S2=2/16=1/8
S3=5/15=1/3
S4=4/20=1/5
Deepest slope S3 aka C
What are we factoring i don't get it
What is the median of the following salary list $32,019, $21,971, $27,512, $43,702, $38,860, $25,997
mylen [45]
Answer:
$29765.50
Step-by-step explanation:
Median is the middlemost value (or the average of the 2 middlemost value if there are even number of values) in an ascending order.
First, arrange the salary list in an ascending order.
$21,971 , $25,997 , $27,512 , $32,019 , $38,860 , $43,702
As we can see here, there are even number of values (6), so the median will be the average of the 2 middlemost value, which will be:
Median = 
= 
= $29765.50