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Fifteen items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following
Find P(2).
Find P(No more than 1).
Find the probability that no one is in line.
Find the probability that at least three people are in line.
Compute the
Compute the standard σX.
If each customer takes 3 minutes to check out: what is the probability that it will take more than 6 minutes for all the customers currently in line to check out?
a.
To find: P(2)
Explanation of Solution
Given information:number of customers atcheckout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x .
the value of P(2) can be read as shown.
We locate x=2 along the x axis and move vertically upward until the height for P(x) is reached. From this point on the line, we move horizontally to the left until the y axis is reached. Reading the P(2) , we find that the P(2)=0.30 .
Therefore,
P(2)=0.30
b.
To find: P( No more than 1)
Explanation of Solution
Given information:number of customers atcheckout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
the values of P(x≤1) can be read as shown.
We locate x=0 and x=1 along the x axis and move vertically upward until the height for P(x) is reached. From this point on the line, we move horizontally to the left until the y axis is reached. Reading the P(0) and P(1) , we find that the P(0)=0.10 and P(1)=0.25 .
Therefore,
P(x≤1)=P(0)+P(1)P(x≤1)=0.10+0.25P(x≤1)=0.35
The P(x≤1) includes both P(0) and P(1) .Because it is for x values not more than 1 .
c.
To find: the probability no one is the line.
Explanation of Solution
Given information:number of customers atcheckout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
the values of P(0) can be read as shown.
We locate x=0 along the x axis and move vertically upward until the height for P(x) is reached. From this point on the line, we move horizontally to the left until the y axis is reached. Reading the P(0) , we find that the P(0)=0.10 .
P(0)=0.10
Therefore,the probability no one is the line is given by P(0) .
d.
To find: the probability at least three people are in the line
Explanation of Solution
Given information:number of customers atcheckout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
the values of P(x≥3) can be read as shown.
We locate x=3 , x=4 , x=5 along the x axis and move vertically upward until the height for P(x) is reached. From this point on the line, we move horizontally to the left until the y axis is reached. Reading the P(3),P(4),P(5) , we find that the P(3)=0.20 , P(4)=0.10 and P(5)=0.05 .
Therefore,the probability that at least three people are in the line is the addition of P(3)=0.20 , P(4)=0.10 and P(5)=0.05 .
P(x≥3)=P(3)+P(4)+P(5)P(x≥3)=0.20+0.10+0.05P(x≥3)=0.35
e.
To calculate: mean μx
Explanation of Solution
Given information:number of customers atcheckout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
mean μx can be calculated using the formula μx=∑x.P(x) as shown.
Therefore,
μx=∑x.P(x)μx=(0)(0.10)+(1)(0.25)+(2)(0.30)+(3)(0.20)+(4)(0.10)+(5)(0.05)μx=0+0.25+0.60+0.60+0.40+0.25μx=2.1
f.
To calculate: standard deviation σx
Explanation of Solution
Given information: number of customers at checkout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
standard deviation σx can be calculated using the formula σ2x=∑[x2P(x)]−μ2x as shown.
Therefore,
σ2x=∑[x2P(x)]−μ2xσ2x=(0)(0.10)+(1)(0.25)+(4)(0.35)+(9)(0.20)+(16)(0.10)+(25)(0.05)−(2.12)σ2x=0+0.25+1.40+1.80+1.60+1.25−4.41σ2x=6.30−4.41σ2x=1.89σx=1.37477270849σx=1.37
f.
To find: the probability that takes more than six minutes (per customer it is 3 minutes) for all the customers in line to check out
Explanation of Solution
Given information: number of customers at checkout counter is a random variable with the following probability distribution.
x012345P(x)0.100.250.300.200.100.05
Graph:the line graph shows P(x) vs. x
the values of P(x≥2) can be read as shown.
We locate x=2,3,4,5 along the x axis and move vertically upward until the height for P(x) is reached. From this point on the line, we move horizontally to the left until the y axis is reached. Reading the P(2),P(3),P(4),P(5) , we find that the P(2)=0.30,P(3)=0.20,P(4)=0.10,P(5)=0.05
Therefore,the probability that takes more than six minutes (per customer it is 3 minutes) for all the customers in line to check out is,
P(x≥2)=P(2)+P(3)+P(4)+P(5)P(x≥2)=0.30+0.20+0.10+0.05P(x≥2)=0.65
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Chapter 6 Solutions
Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
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