Answer:
a. Yes
b. 0.35
c. 0.5
d.
E(x)=34.25
V(x)=73.1875
SD(x)=8.555
Step-by-step explanation:
a.
x 20 25 30 35 40 45 50
f(x) 0.05 0.2 0.25 0.15 0.15 0.1 0.1
This will be valid probability distribution if
1. All probabilities are between 0 and 1 inclusive.
2. Sum of probabilities must be 1.
Condition 1 is satisfied in this distribution. Now, we will check second condition
sum(f(x))=0.05+0.2+0.25+0.15+0.15+0.1+0.1=1.
As the both conditions are satisfied, thus the given distribution is a valid probability distribution.
b.
P(x≥40)=P(X=40)+P(X=45)+P(X=50)
P(x≥40)=0.15+0.1+0.1
P(x≥40)=0.35
So, the probability that Backens and Hayes LLC will obtain 40 or more new clients is 0.35.
c.
P(x<35)=P(X=20)+P(X=25)+P(X=30)
P(x<35)=0.05+0.2+0.25
P(x<35)=0.5
So, the probability that Backens and Hayes LLC will obtain fewer than 35 new clients is 0.5.
d.
Expected value of x=E(x)= sum[x*f(x)]
E(x)= 20*0.05+25*0.2+30*0.25+35*0.15+40*0.15+45*0.1+50*0.1
E(x)=1+5+7.5+5.25+6+4.5+5
E(x)=34.25
Variance of x=V(x)=sum[x²*f(x)]-(sum[x*f(x)])²
sum[x²*f(x)]= 20²*0.05 +25²*0.2 +30²*0.25 +35²*0.15 +40²*0.15 +45²*0.1 +50²*0.1
sum[x²*f(x)]=20+1255+225+183.75+240+202.5+50
sum[x²*f(x)]=1246.25
V(x)=1246.25-(34.25)²
V(x)=73.1875
Standard deviation of x=SD(x)=√V(x)
SD(x)=8.555
