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aliya0001 [1]
3 years ago
5

HELP!!

Mathematics
2 answers:
romanna [79]3 years ago
8 0

Answer:

Step-by-step explanation:

1) P(Z<2.87) = 0.5+P(0<z<2.87)

= 0.9979

Option a

2) P(Z>-0.87) = 0.5+P(-0.87<z<0)

= 0.8078

3) X - scores on a law school ent exam

x bar = 250 and sigma = 50

P(X>284) = P(z>\frac{284-250}{50} =0.68)

=0.5-02517=0.2483

No of students = 320(0.2483) = 79.456

=79

4) X - length of new born N(49.2, 1.8)

N =2000

P(X>51) = P(Z>0.44)

=0.33

No of girls = 2000(0.33) = 660

tamaranim1 [39]3 years ago
6 0
Question 1: <span>The answer is D. which it ended up being <span>0.9979
</span>
Question 2: </span>The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousand

The expression P(z > -0.87) represents the area under the standard normal curve above a given value of z. What is P(z > -0.87)? Express your answer as a decimal to the nearest ten thousandth (four decimal places). So being that rounding it off would mean your answer would be = ?

Question 3: <span>Assume that the test scores from a college admissions test are normally distributed, with a mean of 450 and a standard deviation of 100. 
a. What percentage of the people taking the test score between 400 and 500?
b. Suppose someone receives a score of 630. What percentage of the people taking the test score better? What percentage score worse?
c. A university will not admit a student who does not score in the upper 25% of those taking the test regardless of other criteria. What score is necessary to be considered for admission? </span>z = 600-450 /100 = .5 NORMSDIST(0.5) = .691462<span><span> z = 400-450 /100 = -.5 NORMSDIST(-0.5) = .30854

P( -.5 < z <.5) = .691462 - .30854 = .3829 Or 38.29%
Receiving score of 630:
z = 630-450 /100 = 1.8 NORMSDIST(1.8) = .9641
96.41% score less and 3.59 % score better
upper 25%
z = NORMSINV(0.75)= .6745
.6745 *100 + 450 = 517 Would need score >517 to be considered for admissions </span><span>
Question 4: </span>The z-score for 45cm is found as follows:</span>
Reference to a normal distribution table, gives the cumulative probability as 0.0099.
<span>Therefore about 1% of newborn girls will be 45cm or shorter.</span>
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