The volume of a rectangular prism is represented by the function x

Consider the probability that greater than 26 out of 124 software users will call technical support. Assume the probability that

Mekhanik [1.2K]

Answer:

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

100% probability that greater than 26 out of 124 software users will call technical support.

Step-by-step explanation:

Test if the normal curve can be used as an approximation to the binomial probability by verifying the necessary conditions.

It is needed that:

$np \geq 10$ and $n(1-p) \geq 10$

Binomial probability 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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Out of 124 software users

This means that $n = 124$

Assume the probability that a given software user will call technical support is 97%.

This means that $p = 0.97$

Conditions:

$np = 124*0.97 = 120.28 \geq 10$

$n(1-p) = 124*0.03 = 3.72 < 10$

Since $n(1-p) = 3.72 < 10$ , the normal curve cannot be used as an approximation to the binomial probability.

Consider the probability that greater than 26 out of 124 software users will call technical support.

The lowest possible probability of those is 27, so, if it is 0, since it is considerably below the mean, 100% probability of being greater. We have that:

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

$P(X = 27) = C_{124,27}.(0.97)^{27}.(0.03)^{97} = 0$

1 - 0 = 1

100% probability that greater than 26 out of 124 software users will call technical support.

3 0

3 years ago

Z=32+41.9i

n200080 [17]

A. Re(z)= 32 and Im(z)= 41.9

The real part of a complex number z=a+bi is ‘a’, and the imaginary part is ‘b’

6 0

3 years ago

Which number is not a common factor of 32 and 48

Colt1911 [192]

5, 3, 7, 9 are all non-common factors

8 0

3 years ago

Find the GCF 0f 12 and 18

melisa1 [442]

Answer:

6

Step-by-step explanation:

Using Euclid's algorithm, we divide the larger by the smaller. If the remainder is zero, the divisor is the GCF. Otherwise, we replace the larger with the remainder and repeat.

18 ÷ 12 = 1 r 6

12 ÷ 6 = 2 r 0 . . . . the GCF is 6

__

You can also factor the numbers and see what the common factors are.

18 = 2·3·3

12 = 2·2·3

The common factors are 2·3 = 6.

In the factorizations, we see 2 to powers of 1 and 2, and we see 3 to powers of 1 and 2. The GCF is the product of the common factors to their lowest powers: (2^1)(3^1) = (2)(3) = 6

8 0

2 years ago

GUYS PLEASE HELPP PLSPLSOLSOAKGDJXG

ladessa [460]

Answer:

Let S = starting point

Let ST = length of tunnel

SR = 1414 m

RT = 2236 m

Set up:

ST^2 + SR^2 = RT^2

ST^2 + (1414)^2 = (2236)

Solve for RT to find your answer.

Step-by-step explanation:

5 0

2 years ago