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AVprozaik [17]
3 years ago
14

At the beginning of the month, Sheryl had a balance of $120 in her checking account. She then made a purchase for $80, using her

checking account. How much money will Sheryl need to deposit if she wants her account balance to be the same as it was at the beginning of the month?
Mathematics
1 answer:
kolezko [41]3 years ago
4 0

Answer:

$80

Step-by-step explanation:

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Read 2 more answers
An automobile manufacturer finds that 1 in every 2500 automobiles produced has a particular manufacturing defect. ​(a) Use a bin
Advocard [28]

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a) 0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

b) 0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

Step-by-step explanation:

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

To use the Poisson approximation for the binomial, we have that:

\mu = np

1 in every 2500 automobiles produced has a particular manufacturing defect.

This means that p = \frac{1}{2500} = 0.0004

a) Use a binomial distribution to find the probability of finding 4 cars with the defect in a random sample of 7000 cars.

This is P(X = 4) when n = 7000. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{7000,4}.(0.0004)^{4}.(0.9996)^{6996} = 0.1558

0.1558 = 15.58% probability of finding 4 cars with the defect in a random sample of 7000 cars.

(b) The Poisson distribution can be used to approximate the binomial distribution for large values of n and small values of p.

Using the approximation:

\mu = np = 7000*0.0004 = 2.8. So

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 4) = \frac{e^{-2.8}*(2.8)^{4}}{(4)!} = 0.1557

0.1557 = 15.57% probability of finding 4 cars with the defect in a random sample of 7000 cars. These probabilities are very close, which means that the approximation works.

6 0
3 years ago
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