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1 year ago

(b) A department store has 7,000 charge accounts. The comptroller takes a random sample of 36 of

Mathematics

1 answer:

galben [10]1 year ago

4 0

Answer:

a.0.8664

b. 0.23753

c. 0.15866

Step-by-step explanation:

The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between

a. N164.50 and N185.50?

b. greater than N180.00?

c. less than N168.00?

We solve the above question using z score formula

z = (x-μ)/σ/√n where

x is the raw score,

μ is the population mean = N175

σ is the population standard deviation = N42

n is random number of sample = 36

a. Between N164.50 and N185.50?

For x = N 164.50

z = 164.50 - 175/42 /√36

z = -1.5

Probability value from Z-Table:

P(x = 164.50) = 0.066807

For x = N185.50

z = 185.50 - 175/42 /√36

z =1.5

Probability value from Z-Table:

P(x=185.50) = 0.93319

Hence:

P(x = 185.50) - P(x =164.50)

= 0.93319 - 0.066807

= 0.866383

Approximately = 0.8664

b. greater than N180.00?

x > N 180

Hence:

z = 180 - 175/42 /√36

z = 5/42/6

z = 5/7

= 0.71429

Probability value from Z-Table:

P(x<180) = 0.76247

P(x>180) = 1 - P(x<180) = 0.23753

c. less than N168.00?

x < N168.

z = 168 - 175/42 /√36

z = -7/42/6

z = -7/7

z = -1

Probability value from Z-Table:

P(x<168) = 0.15866

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Let Y denote a geometric random variable with probability of success p. a Show that for a positive integer a, P(Y > a) = qa .

lakkis [162]

Answer:

a) For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

b) $P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this using independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

c) For this case we use independent identical and with the same distribution experiments.

And the result for part b makes sense since we are interest in find the probability that the random variable of interest would be higher than an specified value given another condition with a value lower or equal.

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

If we define the random of variable Y we know that:

$Y\sim Geo (1-p)$

Part a

For this case we can find the cumulative distribution function first:

$F(k) = P(Y \leq k) = \sum_{k'=1}^k P(Y =k')= \sum_{k'=1}^k p(1-p)^{k'-1}= 1-(1-p)^k$

So then by the complement rule we have this:

$P(Y>a) = 1-F(a)= 1- [1-(1-p)^a]= 1-1 +(1-p)^a = (1-p)^a = q^a$

Part b

For this case we can use the result from part a to conclude that:

$P(Y>a)= q^a$

$P(Y>b) = q^b$

So then we have this assuming independence:

$P(Y> a+b) = q^{a+b}$

We want to find the following probability:

$P(Y> a+b |Y>a)$

Using the definition of conditional probability we got:

$P(Y> a+b |Y>a)= \frac{P(Y> a+b \cap Y>a)}{P(Y>a)} = \frac{P(Y>a+b)}{P(Y>a)} = \frac{q^{a+b}}{q^a} = q^b = P(Y>b)$

And we see that if a = 2 and b=5 we have:

$P(Y> 2+5 | Y>2) = P(Y>5)$

Part c

For this case we use independent identical and with the same distribution experiments.

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2 years ago

Ashton surveyed some of the employees at his company about their cell phone habits. From the data, he concluded that most employ

SOVA2 [1]

It can't be A. since if you only look at managers, you are missing all the sales executives.

It may be C. this option is more random but doesn't guarantee that you will represent both groups of employee's. Also, each time you would conduct the survey, you will receive the exact same results since it is the same people.

It isn't D. for the exact same reason as A. but you're missing managers now.

Therefore the answer is B. Some managers and some sales executives selected at random. This way you get a sample from both categories, and within those groups, it is randomly selected.

I hope this helps!

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1 year ago

Perry currently has $180 dollars and is spending $6 per day. Nick has no money but is earning $8 each day. After how many days w

Elan Coil [88]

12.857 days

Step-by-step explanation:

Since the number of days is passing the same for both Nick and Perry we can assign the unknown value (no. Of days) the same denotation of x

Therefore;

180 – 6x = 0 + 8x

180 = 8x + 6x

180 = 14x

X = 12.857

Learn More:

For more on calculating unknown values check out;

brainly.com/question/10578482

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2 years ago

Kiersten runs a website that helps people learn programming. Every month, Kiersten receives a subscription fee of \$10$10dollar

NemiM [27]

700009000720067261836

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1 year ago

Read 2 more answers

In the transmission of digital information, the probability that a bit has high, moderate, and low distortion is 0.02, 0.07, and

Leviafan [203]

Answer:

A. (Table Attached)

B. (See Step 3)

C. 0.06 (See Step 4)

D. NOT independent (See Step 5)

Step-by-step explanation:

Name the probabilities:

p₁ = 0.02, p₂ = 0.07, p₃ = 0.91

q₁ = 1-p₁ = 0.98 , q₂ = 1-p₂ = 0.93 , q₃ = 0.09

Let X and Y be the number of bits with high and moderate distortion out of three.

The function will follow multinomial distribution:

$f_{XY}(x,y) = P(X=x, Y=y) = \frac{3!}{x!y!(3-x-y)!} (p_1^x)(p_2^y)(p_3^{3-x-y})$

Substitute the values and make a table.

TABLE IN ATTACHMENT

We calculate marginal distribution by:

$P (X=x)=$ <em>∑</em> $P(X=x,Y=y)$

$fx(x)$ can be found by adding all the probabilities in each row for different value of X

For X=0 , ∑P = 0.94157441

For X=1 , ∑P = 0.057624

For X=2 , ∑P = 0.001176

For X=3 , ∑P =0.000008

The mathematic expectation E is the sum of product of each possibility with its probabiity.

$E(X)=$ <em>∑ </em> $xP(X=x)$

Find E(X):

$E(X)= (0*0.9415744)+(1*0.057624)+(2*0.001176)+(3*0.000008)$

$E(X)=0.06$

Condition probability states:

$P(A|B)=\frac{P(A,B)}{P(B)}$

It can also be written as:

$f_{Y|X=1}(y)=\frac{f_{XY}(1,y)}{f_x(1)}$

Where $f_x(1)\\$ = 0.057624

Calculate the quotient:

$Y|_{x=1}$ = 0 , $f_{Y|_X=1$ = 0.862245

$Y|_{x=1}$ = 1 , $f_{Y|_X=1$ = 0.132653

$Y|_{x=1}$ = 2 , $f_{Y|_X=1$ = 0.000510

$Y|_{x=1}$ = 3 , $f_{Y|_X=1$ = 0

Find the dependency:

$f_{XY}(y)=f_X(x)f_Y(y)$

We found that

$f_{Y|_X=1$ = 0.862245

Calculate $f_Y(1)$ from summing the column from the table

$f_Y(1)=0.17428341+0.007644+0.000084\\f_Y(1)=0.18201141$

Which are not equal.

Conclusion:

X and Y are NOT Independent

7 0

2 years ago