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3 years ago

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Mamady is planning a trip. He will stay six nights at a hotel that charges $51.25 a night. What is a good estimate of the total

amount he will pay for the hotel? 250, 300, 360, 50.

1 answer:

frez [133]3 years ago

5 0

Answer:

Answer is 300

Step-by-step explanation:

because 51.25x6=307.5 and the closest estimate is 300.

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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.

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