What is the area of this figure?

Consider a binomial experiment with 15 trials and probability 0.35 of success on a single trial.

DedPeter [7]

Answer:

a

$P(X = 10 ) = 0.0096$

b

$P(X = 10 ) = 0.0085$

c

Option A is correct

Step-by-step explanation:

From the question we are told that

The sample size is n = 15

The probability of success is $p = 0.35$

The number of success we are considering is r = 10

Now the probability of failure is mathematically evaluated as

$q = 1- p$

substituting value

$q = 1- 0.35$

$q = 0.65$

Now using the binomial distribution to find the probability of exactly 10 successes we have that

$P(X = r ) = [\left n } \atop {r}} \right. ] * p^r * q^{n- r}$

substituting values

$P(X = 10 ) = [\left 15 } \atop {10}} \right. ] * p^{10}* q^{15- 10}$

Where $[\left 15 } \atop {10}} \right. ]$ mean 15 combination 10 which is evaluated with a calculator to obtain

$[\left 15 } \atop {10}} \right. ] = 3003$

So

$P(X = 10 ) = 3003 * 0.35 ^{10}* 0.65^{15- 10}$

$P(X = 10 ) = 0.0096$

Now using the normal distribution to approximate the probability of exactly 10 successes, we have that

$P(X = r ) = P( r < X < r )$

Applying continuity correction

$P(X = r ) = P( r -0.5 < X < r +0.5)$

substituting values

$P(X = 10) = P( 10-0.5 < X < 10+0.5)$

$P(X = 10 ) = P( 9.5 < X < 10.5)$

Standardizing

$P(X = r ) = P( \frac{9.5 - \mu }{\sigma } < \frac{X - \mu }{\sigma } < \frac{10.5 - \mu}{\sigma } )$

The where $\mu$ is the mean which is mathematically represented as

$\mu = n * p$

substituting values

$\mu = 15 * 0.35$

$\mu = 5.25$

The standard deviation is evaluated as

$\sigma = \sqrt{n * p * q }$

substituting values

$\sigma = \sqrt{15 * 0.35 * 0.65 }$

$\sigma = 1.8473$

Thus

$P(X = 10 ) = P( \frac{9.5 - 5.25 }{1.8473 } < \frac{X - 5.25 }{1.8473 } < \frac{10.5 - 5.25}{1.8473 } )$

$P(X = 10 ) = P( 2.30 < Z < 2.842 )$

$P(X = 10 ) = P(Z < 2.842 ) - P(Z < 2.30 )$

From the normal distribution table we obtain the $P(Z < 2.841)$ as

$P(Z < 2.841) = 0.99775$

And the $P(Z < 2.30)$

$P(Z < 2.30) = 0.98928$

There value can also be obtained from a probability of z calculator at (Calculator dot net website)

So

$P(X = 10) = 0.99775 - 0.98928$

$P(X = 10 ) = 0.0085$

Looking at the calculated values for question a and b we see that the values are fairly different.

8 0

4 years ago

Mark put his dog on a diet. The dog's total weight change for the first 2 weeks was −3/4 pound. The dog lost the same amount of

prohojiy [21]

About 1/3 of a pound

4 0

3 years ago

Read 2 more answers

According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the

FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let $P(t)$ be the amount of $^{235}U$ and $Q(t)$ be the amount of $^{238}U$ after $t$ years.

Then, we obtain two differential equations

$\frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q$

where $k_1$ and $k_2$ are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

$\frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt$

Now, the variables are separated, $P$ and $Q$ appear only on the left, and $t$ appears only on the right, so that we can integrate both sides.

$\int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt$

which yields

$\ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2$ ,

where $c_1$ and $c_2$ are constants of integration.

By taking exponents, we obtain

$e^{\ln |P|} = e^{-k_1t + c_1} \quad e^{\ln |Q|} = e^{-k_12t + c_2}$

Hence,

$P = C_1e^{-k_1t} \quad Q = C_2e^{-k_2t}$ ,

where $C_1 := \pm e^{c_1}$ and $C_2 := \pm e^{c_2}$ .

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

$P(0) = Q(0) = C$

Substituting 0 for $P$ in the general solution gives

$C = P(0) = C_1 e^0 \implies C= C_1$

Similarly, we obtain $C = C_2$ and

$P = Ce^{-k_1t} \quad Q = Ce^{-k_2t}$

The relation between the decay constant $k$ and the half-life is given by

$\tau = \frac{\ln 2}{k}$

We can use this fact to determine the numeric values of the decay constants $k_1$ and $k_2$ . Thus,

$4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}$

and

$7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}$

Therefore,

$P = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}$

We have that

$\frac{P(t)}{Q(t)} = 137.7$

Hence,

$\frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7$

Solving for $t$ yields $t \approx 6 \times 10^9$ , which means that the age of the universe is about 6 billion years.

5 0

3 years ago

Whats the solution to this equation? 3x-9=12

Nutka1998 [239]

Answer:

x=7

Step-by-step explanation:

3x-9=12

add 9 to both sides

3x=21

divide both sides by 3

x=7

6 0

3 years ago

Read 2 more answers

Develop an estimated multiple linear regression model that could be used to predict the alumni giving rate using the graduation

madam [21]

The answer to this question is false that's the answer

4 0

4 years ago