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zysi [14]
4 years ago
5

IQ is normally distributed with a mean of 100 and a standard deviation of 15. a) Suppose one individual is randomly chosen. Fin

d the probability that this person has an IQ greater than 95. Write your answer in percent form. Round to the nearest tenth of a percent. P (IQ greater than 95) = ________ % b) Suppose one individual is randomly chosen. Find the probability that this person has an IQ less than 125. Write your answer in percent form. Round to the nearest tenth of a percent. P (IQ less than 125) = ___________% c) In a sample of 500 people, how many people would have an IQ less than 110? d) In a sample of 500 people, how many people would have an IQ greater than 140? people
Mathematics
1 answer:
nasty-shy [4]4 years ago
3 0

Answer:

Step-by-step explanation:

Let X be the IQ

IQ is normally distributed with a mean of 100 and a standard deviation of 15.

a) the probability that this person has an IQ greater than 95

=P(X\geq 95)\\=P(Z\geq \frac{95-100}{15} \\=P(Z\geq -0.33)\\=0.1293+0.5\\=0.6293

=62.93%

b) the probability that this person has an IQ less than 125

=P(X\leq 125) =P(X\leq 1.67)\\=0.5-0.4525\\=0.0475

=4.75%

c) Sample size =500

P(X

No of people = 0.2486(500) =124.3

d) P(x>140)\\= P(Z>2.6)\\=0.5-0.4953\\=0.0047

No of persons = 0.0047(500) = 2.35

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zloy xaker [14]

Part A: y=-\frac{5}{4} x+\frac{35}{2} is the equation of the line.

Part B: Anika worked 17.5 hours on day 0.

Part C: The setup time decreases with 1 hour and 15 min per day.

Explanation:

Part A: Let us find the equation of the line using the points (2,15) and (6,10)

The equation of the line is given by,

y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot\left(x-x_{1}\right)

Substituting, we get,

y-15=\frac{10-15}{6-2} (x-2)

Simplifying, we have,

y-15=-\frac{5}{4} (x-2)

y-15=-\frac{5}{4} x+\frac{5}{2}

Subtracting both sides by 15, we get,

y=-\frac{5}{4} x+\frac{35}{2}

Hence, the equation of the line is y=-\frac{5}{4} x+\frac{35}{2}

Part B: To determine the number of hours Anika worked on day 0, let us substitute x = 0 in the equation of the line.

Thus, we get,

y=\frac{35}{2}\\y=17.5

Hence, Anika worked 17.5 hours on day 0.

Part C: The setup time decrease per day can be determined using the slope.

The formula for slope is given by

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Substituting , we have,

m=\frac{10-15}{6-2}=-\frac{5}{4}

Converting the fraction into hours and minutes, we get,

-\frac{5}{4}\times60=-75 min

Since, 1 hour = 60 min

Subtracting 60 and 75 = 15 min

Thus, the setup time decreases with 1 hour and 15 min per day.

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It costs $3.00 for 45 gallons of detergent. What is the cost per gallon of detergent? *
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0,06

Step-by-step explanation:

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if i helped please give brainliest and a thanks !!
3 0
3 years ago
One of ten different prizes was randomly put into each box of a cereal. If a family decided to buy this cereal until it obtained
Reptile [31]

Answer:

29.29

Step-by-step explanation:

The first trial among a distribution of independent trials with a constant success probability follows a geometric probability.

The expected value of a negative binomial is the reciprocal of its success probability <em>p.</em>

<em>E(X)=</em>1/<em>p</em>

At the first draw, we only selected one of the 10 prizes and so 1 draw gives a success

<em>E(X)=</em>1/<em>p</em>

<em>E(X1)</em>=1

After the first prize is drawn, we are interested in the first time (success), that we draw another prize that is different (<em>p</em>=9/10)

E(X2)=1/9÷10

E(X2)= 10/9

After the first prizes have being drawn, we now interested in the first time success that we draw a different prize from the first 2

E(X3)=1/8÷10

E(X3)=10/8

E(X3)=5/4

After the first three prizes have been drawn, we are now interested in the success of a first time draw for the 4th different prize

E(X4)=1/7÷10

E(X4)=10/7

Fifth different prize

E(X5)=1/6÷10

E(X5)=10/6

E(X5)=5/3

Sixth different prize

E(X6)=1/5÷10

E(X6)=10/5

E(X6)=2

Seventh different prize

E(X7)=1/4÷10

E(X7)=10/4

E(X7)=5/2

Eighth different prize

E(X8)=1/3÷10

E(X8)=10/3

Ninth different prize

E(X9)=1/2÷10

E(X9)=10/2

E(X9)=5

After the 9 different prizes have been drawn, we are interested in the first time we will draw the tenth different prize

E(X10)=1/1÷10

E(X10)=10

Add up all corresponding values;

<em>E(X)=</em>E(X1)+E(X2)+E(X3)+E(X4)+E(X5)+E(X6)+E(X7)+E(X8)+E(X9)+E(X10)

<em>E(X)=</em>1+10/9+5/4+10/7+5/3+2+5/2+10/3+5+10

<em>E(X)</em>= 29.29

Note: Those values in <em>E(X), </em>should be written the way it will in standard algebra ie they should be small.

4 0
3 years ago
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