A. I, III, II b. II, I, III c. II, III, I d. III, I, II

To be considered as a menu item for Gloria's new restaurant an item must have scored in the top 15% by the food critics in the a

vladimir1956 [14]

Answer:

Lowest score needed=4.92

Step-by-step explanation:

Using the standard normal distribution table we find the value of standard normal deviate corresponding to area of 15%

For area of 15% we have Z= -1.04

Thus we have

$Z=\frac{X-\overline{X}}{\sigma }\\\\\therefore X=\sigma Z+\overline{X}\\\\$

Applying values we get

$X=-1.04\times 2+7\\\\X=4.92$

Thus lowest score needed = 4.92

3 0

3 years ago

A rectangle has an area of 20 square units what can be said if it's height and width

Butoxors [25]

The height could be 2 and the width 10 or the height could be 10 and the width could be 2.

4 0

2 years ago

Read 2 more answers

The table shows the total time it took Samir to read 0,1,2, and 3 pages of the book. The table also list this information as ord

cluponka [151]

Ok, but where is the table???

8 0

3 years ago

I need help with this please

Levart [38]

Yu can you think how many triangles fit in that parallelogram? The answer is six. four in the central rectangle and two at the ends, then the area of the parallelogram will be six times greater than that of the triangle in that graph. Good luck! M.

7 0

3 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of trials before q successes is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

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

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

One person, thus $q = 1$ .

The expected value is:

$E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4$

The variance is:

$V = \frac{0.8}{0.04} = 20$

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0

1 year ago