Question

17) Not everyone pays the same price for the same model of a car. For a particular new car the mean price is $18,750 with a standard deviation of $690.   ROUND VALUES CORRECTLY TO DOLLAR.

a) Make a sketch of the normal distribution, labeling the center and ± 1, 2, 3SD intervals with the values.

b) If a car is picked at random, what is the probability that the car price is between $17,800 and $19,100?

c) A car in the top 12% will have a price of at least how much? (To nearest dollar)

d) We take many samples of 30 cars of this model. Find the standard deviation of the sample to nearest dollar. (no sentence)

e) Sketch the sampling distribution if we took many samples of 30 of these models of car. (no sentence)

f) What is the probability that a sample of 30 cars, this model, will have a mean price cost of $18,400 or less?

g) A sample of 30 cars, this model, that has an average price in the top car in the bottom 5% will average price at most how much? (To nearest dollar)

Answer 1

b)

µ =    18750                              
σ =    690                              
we need to calculate probability for ,                                  
P (   17800   < X <   19100   )                  
=P( (17800-18750)/690 < (X-µ)/σ < (19100-18750)/690 )                                  
                                  
P (    -1.377   < Z <    0.507   )                   
= P ( Z <    0.507   ) - P ( Z <   -1.377   ) =    0.6940   -    0.0843   =    0.6097

.............

c)

µ=   18750                  
σ =    690                  
proportion=   0.88                  
                      
Z value at    0.88   =   1.17   (excel formula =NORMSINV(   0.88   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   1.17   *   690   +   18750  
X   =   19561   (answer)          
......

d)

σ =    690
n=   30
std dev=(σ/√n) = 126

.................

f)

µ =    18750                                  
σ =    690                                  
n=   30                                  
                                      
X =   18400                                  
                                      
Z =   (X - µ )/(σ/√n) = (   18400   -   18750.00   ) / (   690.00   / √   30   ) =   -2.778
                                      
P(X ≤   18400   ) = P(Z ≤   -2.778   ) =   0.0027                 
.........

g)

µ =    18750                              
σ =    690                              
n=   30                              
proportion=   0.9500                              
                                  
Z value at    0.95   =   1.645   (excel formula =NORMSINV(   0.95   ) )          
z=(x-µ)/(σ/√n)                                  
so, X=z * σ/√n +µ=   1.645   *   690   / √    30   +   18750.00   = 18957
..................

THANKS

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