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IrinaK [193]
2 years ago
12

(2) The scores obtained by a class of

Mathematics
1 answer:
SVETLANKA909090 [29]2 years ago
5 0

Answer:

Below.

Step-by-step explanation:

2)

(a) Prob(exactly 6 marks) =  2/10 = 1/5

(b) Prob(At least 6 marks) = 5/10 = 1/2.

3)

Prob ( Queen of hearts) = 1/52

4)

Probability of a 6 in one throw = 1/6.

So expected number of 6's in 300 throws

= 1/6 * 300

= 50.

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The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir
Kazeer [188]

Answer:

(1) (c) <u>5.30 years</u>.

(2) (b) <u>0.289</u>.

(3) (b) <u>0.80</u>.

(4) (d) <u>0.50</u>.

(5) (a) <u>5.25 years</u>.

Step-by-step explanation:

Let <em>X</em> = age of the children in kindergarten on the first day of school.

The random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 4.8 years and <em>b</em> = 5.8 years.

The probability density function function of <em>X</em> is:

f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a

(1)

The expected value of a Uniform random variable is:

E(X)=\frac{1}{2}(a+b)

Compute the mean of <em>X</em> as follows:

E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3

Thus, the  mean of the distribution is (c) <u>5.30 years</u>.

(2)

The standard deviation of a Uniform random variable is:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289

Thus, the standard deviation of the distribution is (b) <u>0.289</u>.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\

                =\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8

Thus, the probability that a randomly selected child is older than 5 years old is (b) <u>0.80</u>.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

P(5.2

                            =\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) <u>0.50</u>.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of <em>x</em> as follows:

   P (X < x) = 0.45

\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45

        \int\limits^{x}_{4.8} {1}\, dx=0.45

           [x]^{x}_{4.8}=0.45

       x-4.8=0.45\\

                x=0.45+4.8\\x=5.25

Thus, the age of the child at the 45th percentile is (a) <u>5.25 years</u>.

6 0
3 years ago
(-0.3) + 0.9 find the sum
IceJOKER [234]

Answer: (-0.3) + 0.9= 0.6


Step-by-step explanation:


3 0
4 years ago
Read 2 more answers
Help me plz thank you
DIA [1.3K]

Answer:C

Step-by-step explanation: 100 x 23/100=23

8 0
3 years ago
What is 625 in exponential form
Anit [1.1K]
5 x 5 x 5 x 5 = 625,

3 0
4 years ago
Read 2 more answers
there are 378 million children who probably believe in Santa Claus. but some have been naughty so they don't get presents. howev
Anna11 [10]
Depends on how much time he has
if he travels against the rotation of the earth, he has more time
the answer would be number of stops/time (minutes)
number of stops=91.8million

if we were to assume he did it at night (from 6 to 6) disregarding complecations like roration and bathroom breaks, 12 hours=720 minues
91.8million/720=127500 stops per minute
8 0
3 years ago
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