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schepotkina [342]
3 years ago
15

A true-false quiz with 10 questions was given to a statistics class. Following is the probability distribution for the score of

a randomly chosen student. Find the mean score and interpret the result. Round the answers to two decimal places as needed.

Mathematics
2 answers:
polet [3.4K]3 years ago
8 0

Answer:

hi your question is incomplete here is the complete question attached to the answer is the probability distribution of the randomly chosen student

answer : 7.5

Step-by-step explanation:

The mean score of the probability distribution

= summation of (x * P(x) )

= ( 5* 0.05 ) + ( 6*0.16 ) + ( 7 * 0.32 ) + ( 8 * 0.26 ) + ( 9 * 0.13 ) + ( 10* 0.08)

= 0.25 + 0.96 + 2.24 + 2.08 + 1.17 + 0.8

= 7.5  the mean score is 7.5

This result means that the mean if more students were given the True -false  quiz with 10 questions the mean score would still be 7.5

disa [49]3 years ago
5 0

Answer:

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

X         5       6         7       8        9        10

P(X)   0.05   0.15  0.33  0.28   0.12   0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:  

E(X)=\sum_{i=1}^n X_i P(X_i)  

And if we use the values obtained we got:  

E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48  

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :  

E(X^2)=\sum_{i=1}^n X^2_i P(X_i)  

And using the formula we got:  

E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46  

Then we can find the variance with the following formula:  

Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096  

And then the standard deviation would be given by:  

Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229  

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Is -2 a solution of the inequality -2x + 5 &gt; 9?<br> A. yes<br> | B. no
kramer

Answer:B

Step-by-step explanation:

-2x+5>9

-2x>4

-x>2

x<-2

3 0
3 years ago
Find a recurrence relation for the amount of money in a savings account after n years if the interest rate is 6 percent and $50
Korvikt [17]

Answer:

a_n = a_{n-1} (1.06) + 50

Step-by-step explanation:

Suppose, a_0 is initial amount in the saving account,

Here, the annual interest rate is 6% and additional amount in each year is $ 50,

So, the amount after one year,

a_1 = a_0 + 6\%\text{ of }a_0 + 50 = a_0 + 0.06a_0 + 50 = a_0(1.06) + 50

Amount after 2 years,

a_2 = a_1 + 6\%\text{ of }a_1 + 50 = a_1(1.06) + 50

Amount after 3 years,

a_3 = a_2 + 6\%\text{ of }a_2 + 50 = a_2(1.06) + 50

................................., so on....

Hence, by following the pattern,

The amount after n years,

a_n = a_{n-1} (1.06) + 50

Which is the required recurrence relation for the amount of money in a savings account

3 0
3 years ago
Value of x in this? Having trouble with this problem.
Solnce55 [7]
Answer:

I think it’s 105

Step by step explanation:

I hope this helped in some kind of way.
6 0
3 years ago
I need help with #1
kolezko [41]
D. Because it corresponds to the angle to the left
5 0
3 years ago
When a breeding group of animals is introduced into a restricted area such as a wildlife reserve, the population can be expected
jasenka [17]

Answer:

A. Initially, there were 12 deer.

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. After 15 years, there will be 410 deer.

D. The deer population incresed by 30 specimens.

Step-by-step explanation:

N=\frac{12.36}{0.03+0.55^t}

The amount of deer that were initally in the reserve corresponds to the value of N when t=0

N=\frac{12.36}{0.33+0.55^0}

N=\frac{12.36}{0.03+1} =\frac{12.36}{1.03} = 12

A. Initially, there were 12 deer.

B. N(10)=\frac{12.36}{0.03 + 0.55^t} =\frac{12.36}{0.03 + 0.0025}=\frac{12.36}{y}=380

B. <em>N(10)</em> corresponds to the amount of deer after 10 years since the herd was introducted on the reserve.

C. N(15)=\frac{12.36}{0.03+0.55^15}=\frac{12.36}{0.03 + 0.00013}=\frac{12.36}{0.03013}= 410

C. After 15 years, there will be 410 deer.

D. The variation on the amount of deer from the 10th year to the 15th year is given by the next expression:

ΔN=N(15)-N(10)

ΔN=410 deer - 380 deer

ΔN= 30 deer.

D. The deer population incresed by 30 specimens.

8 0
3 years ago
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