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Molodets [167]
3 years ago
9

Suppose SAT Writing scores are normally distributed with a mean of 498 and a standard deviation of 108. A university plans to aw

ard scholarships to students whose scores are in the top 4%. What is the minimum score required for the scholarship? Round your answer to the nearest whole number, if necessary.

Mathematics
1 answer:
Katen [24]3 years ago
5 0

Answer: 687

Step-by-step explanation:

normally distributed

μ = 498

σ = 108

Top 4%

P(X > a) = 0.04

P(Z > a-498/108) = 0.04

Using the table attached, for the top 4%:

0.50 - 0.04 = 0.46

a-498/108 = 1.75

a-498 = 189

a = 687

The minimum score for the scholarship is 687.

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Given the vectors u= <2, 1> and v= <5, 4>, determine the components of a vector represented by 2u - 3v
kati45 [8]

Answer:

C.) <-11, -10>

Step-by-step explanation:

Let's define how to work with vectors.

For two vectors:

V = <a, b>

W = <c, d>

The product of a scalar k and a vector is given by:

k*V = k*<a, b> = <k*a, k*b>

And the sum (or difference) of two vectors is given by:

V ± W  = <a, b> ± <c, d> = <a ± c, b ± d>

Now that we know this, we can solve the problem.

Here we have the vectors:

u = <2, 1>

v = <5, 4>

then:

2u - 3*v = 2*<2, 1> - 3*<5, 4>

             = <2*2, 2*1> - <3*5, 3*4>

             = <4, 2> - <15, 12>

             = <4 - 15, 2 - 12>

             = < -11, -10>

Then the correct option is C.

5 0
3 years ago
Consider the next 1000 90% CIs for μ that a statistical consultant will obtain for various clients. Suppose the data sets on whi
Ivenika [448]

Answer:

90% CI expects to capture u 90% of time

(a) This means 0.9 * 1000 = 900 intervals will capture u

(b) Here we treat CI as binomial random variable, having probability 0.9 for success

n = 1000

p = 0.9

For this case, applying normal approximation to binomial, we get:

mean = n*p= 900

variance = n*p*(1-p) = 90

std dev = 9.4868

We want to Find : P(890 <= X <= 910) = P( 889.5 < X < 910.5) (integer continuity correction)

We convert to standard normal form, Z ~ N(0,1) by z1 = (x1 - u )/s

so z1 = (889.5 - 900 )/9.4868 = -1.11

so z2 = (910.5 - 900 )/9.4868 = 1.11

P( 889.5 < X < 910.5) = P(z1 < Z < z2) = P( Z < 1.11) - P(Z < -1.11)

= 0.8665 - 0.1335

= 0.733

6 0
3 years ago
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