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3 years ago

The following table contains the probability distribution for the number of traffic accidents daily in a small city:

Mathematics

1 answer:

8_murik_8 [283]3 years ago

8 0

Answer:

a) 2 accidents per day

b) .7071 accidents

Step-by-step explanation:

a) The mean is the sum of each possible number of accidents multiplied by its respective probability. So,

E(x) = (0)(.1) + (1)(.2) + (2)(.45) + (3)(.15) + (4)(.05) + (5)(.05) = 2

b) The variance V(x) = E(x^2) - E(x)^2, and the standard deviation is the square root of the variance. So,

V(x) = E(x^2) - (2)^2

= (0)(.1) + (1)(.2) + (4)(.45) + (9)(105) + (16)(.05) + (25)(.05) - 4 = 0.5

sqrt(0.5) = .7071

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pashok25 [27]

Answer:

Step-by-step explanation:

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3 years ago

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A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$

Hence, 88.51 is the minimum score needed to receive a grade of A.

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3 years ago

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Sedbober [7]

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3 years ago

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4 years ago

Find the smallest positive integer solution to the following system of congruences: x = 0 (mod 5) x = 8 (mod 11) The solution is

igomit [66]

Answer:

The smallest positive integer solution to the given system of congruences is 30.

Step-by-step explanation:

The given system of congruences is

$x=0(mod5)$

$x=8(mod11)$

where, m and n are positive integers.

It means, if the number divided by 5, then remainder is 0 and if the same number is divided by 11, then the remainder is 8. It can be defined as

$x=5m$

$x=11n+8$

$5m\cong 11n+8$

Now, we can say that m>n because m and n are positive integers.

For n=1,

$5m=11(1)+8=19$

$5m=19$

19 is not divisible by 5 so m is not an integer for n=1.

For n=2,

$5m=11(2)+8$

$5m=30$

$m=6$

The value of m is 6 and the value of n is 2. So the smallest positive integer solution to the given system of congruences is

$x=5(6)=30$

Therefore the smallest positive integer solution to the given system of congruences is 30.

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3 years ago