a car dealer sells suvs and sedans at a ratio of 3:2 each month. if the dealer reports selling 95 cars in a month, how many of e
1 answer:
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Answer:
a) 20.16; b) 20.49 and 21.51
Step-by-step explanation:
We use z scores for each of these. The formula for a z score is
.
For part a, we want the 20th percentile; this means we want 20% of the data to be lower than this. We find the value in the cells of the z table that are the closest to 0.20 as we can get; this is 0.2005, which corresponds with a z score of -0.84.
Using this, 21 as the mean and 1 as the standard deviation,
-0.84 = (X-21)/1
-0.84 = X-21
Add 21 to each side:
-0.84+21 = X-21+21
20.16 = X
For part b, we want the middle 39%. This means we want 39/2 = 19.5% above the mean and 19.5% below the mean; this means we want
50-19.5 = 30.5% = 0.305 and
50+19.5 = 69.5% = 0.695.
Looking these values up in the cells of the z table, we find those exact values; 0.305 corresponds with z = -0.51 and 0.695 corresponds with z = 0.51:
-0.51 = (X-21)/1
-0.51 = X-21
Add 21 to each side:
-0.51+21 = X-21+21
20.49 = X
0.51 = (X-21)/1
0.51 = X-21
Add 21 to each side:
0.51+21 = X-21+21
21.51 = X
Answer:
Step-by-step explanation:
Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by
f(x,y) = P((X=x,Y=y) ,
and given that the random variables are independent the joint pmf isbgiven by:
f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)
(b) the required probability is given by considering X=0,1 and Y =0,1
f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }
= [0.1 +0.2 ] × [0.1 + 0.3]
= 0.3 × 0.4
= 0.12
Now we find
fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1
fX(x) = sum{[f(x,y)]}= 0.1+0.3
= 0.4
fY(y) = sum{[f(x.y)]} = 0.1 + 0.2
= 0.3
Since fY(y) × fX(x) = 0.4 × 0.3
= 0.12
Hence f(x,y) = fX(x) .fY(y), the events are independent
(c) the required event is give by: [P(X<=1, PY<=1)]
= P(X<=1) . P(Y<=1)
= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1
= 0.4 × 0.3
= 0.12
(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A
A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]
A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]
A = [0.09] + [ 0.09]
A = 0.18
Pls note the sum of the vertical column X=2 is 0.3
