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oee [108]
3 years ago
7

What proportion of a normal distribution is located between z = 1.00 and z = 1.50?

Mathematics
1 answer:
lesya692 [45]3 years ago
4 0

Answer:

0.092 is the required proportion.

Step-by-step explanation:

We are given the following in the question:

A normal distribution.

We have to find the proportion of data lying between z = 1.00 and z = 1.50.

P(between z = 1.00 and z = 1.50)

P(.001 \leq z \leq 1.50)\\= P(z \leq 1.50) - P(z < 1.00)\\= 0.933 - 0.841 = 0.092 = 9.2\%

Thus, 0.092 is the proportion of data located between z = 1.00 and z = 1.50.

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Answer: 25.6

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Granola costs $6.50 per pound. Assorted nuts cost $9 per pound. How many pounds of granola and nuts are needed to make 9 pounds
bezimeni [28]

Answer:

  • 3.6 pounds of granola and 5.4 pounds of nuts

Step-by-step explanation:

Let the required granola be x pounds, then amount of nuts is 9 - x.

<u>The price of the mixture is as below equation:</u>

  • 6.5x + 9(9 - x) = 8*9
  • 6.5x + 81 - 9x = 72
  • 81 - 2.5x = 72
  • 2.5x = 81 - 72
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<u>Find the amount of nuts:</u>

  • 9 - 3.6 = 5.4
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2 years ago
Write and solve an equation to answer the question.
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3 years ago
-3 + { 1 – 3[20 + 2(-6 – 7)]}
mart [117]

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3 years ago
A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
Readme [11.4K]

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

3 0
3 years ago
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