One year consumers spent an average of $23 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $6. Complete parts (a) through (c) below.a. What is the probability that a randomly selected person spent more than $28?=0.2033
b. What is the probability that a randomly selected person spent between $9 and $21?=0.3608

Answers

Answer 1

Answer:

The probabilities regarding a person spending are given as follows:

a) More than 28: 0.2033 = 20.33%.

b) Between 9 and 21: 0.3608 = 36.08%.

How to obtain probabilities using the normal distribution?

The z-score of a measure X of a variable that has mean symbolized by $\mu$ and standard deviation symbolized by $\sigma$ is obtained by the rule presented as follows:

$Z = (X - \mu)/(\sigma)$

The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.

Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.

The mean and the standard deviation for this problem are given as follows:

$\mu = 23, \sigma = 6$

The probability of a person spending more than 28 is one subtracted by the p-value of Z when X = 28, hence:

Z = (28 - 23)/6

Z = 0.83

Z = 0.83 has a p-value of 0.7967.

1 - 0.7967 = 0.2033 = 20.33%.

The probability of spending between 9 and 21 is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 9, hence:

Z = (21 - 23)/6

Z = -0.33

Z = -0.33 has a p-value of 0.3707.

Z = (9 - 23)/6

Z = -2.33

Z = -2.33 has a p-value of 0.0099.

Hence:

0.3707 - 0.0099 = 0.3608 = 36.08%.

More can be learned about the normal distribution at brainly.com/question/25800303

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0.06 or 0.06666666666 As a fraction

Answers

0.06 = 6/100
When simplified, that's 3/50 .

0.06666666666 = 6,666,666,666/10,000,000,000
When simplified, that's 3,333,333,333/5,000,000,000 .

In 195019501950, the per capita gross domestic product (GDP) of Australia was approximately \$1800$1800dollar sign, 1800. Each year afterwards, the per capita GDP increased by approximately 6.7\%6.7%6, point, 7, percent. Write a function that gives the approximate per capita GDP G(t)G(t)G, left parenthesis, t, right parenthesis of Australia ttt years after 195019501950. Do not enter commas in your answer.

Answers

Answer:

G(t) = 1800(1.067)^t

Step-by-step explanation:

Answer:

G(t) = 1800(1.067)^t

Step-by-step explanation:

To simplify each expression and use positive exponents for m^3n^-6p^0.

Answers

$m^(3n) -6 p^(0)$
= $m^(3n) -6$

$m^3n^-6p^0$

$Use \ Negative \ Power \ Rule$ $x^-^a= (1)/(x^a)$

$m^3* (1n)/(6) p^0$

$Use \ Rule \ of \ Zero$ $x^0=1$

$m^3*1 (1)/(n^6) *1$

$\boxed{\boxed{(m^3)/(n^6)}}$ $Solution$

A factory produces 1000 bottles of pasta sauce in a day. Fifty bottles are selected at random to determine if they weigh the correct amount. The factory rejects bottles that are over or under the correct weight.One bottle is under weight, 46 bottles are the correct weight, and 3 bottles are over weight.

Based on this random sample, approximately how many bottles will the factory reject from the day's production?

A.4

B.20

C.60

D.80

Answers

Answer:

option: D is correct.

Step-by-step explanation:

The random sample is given of 50 bottles which is depicted as:

1 bottle under weight

46 bottles correct weight

3 bottles over weight

As the bottles which are under weight and over weight are rejected.

that means out of 50: 4 bottles are rejected.

Now we have to find out the number of bottles rejected out of the 1000 bottles.

As 1000 is 20 times of 50.

Hence the number of bottles rejected will also get multiplied by 20.

Hence total number of bottles rejected=80.

Hence, option D is correct.

the answer would consiste of c

(sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^

Answers

$(sin\theta+sec\theta)^2+(cos\theta+cosec\theta)^2=(1+sec\theta\ cosec\theta)^2\n\nL=sin^2\theta+2sin\theta\cdot(1)/(cos\thewta)+(1)/(cos^2\theta)+cos^2\theta+2cos\theta\cdot(1)/(sin\theta)+(1)/(sin^2\theta)\n\n=(sin^2\theta+cos^2\theta)+(2sin\theta)/(cos\theta)+(2cos\theta)/(sin\theta)+(1)/(cos^2\theta)+(1)/(sin^2\theta)$

$=1+(2sin^2\theta+2cos^2\theta)/(sin\theta\ cos\theta)+(sin^2\theta+cos^2\theta)/(sin^2\theta\ cos^2\theta)\n\n=1+(2(sin^2\theta+cos^2\theta))/(sin\theta\ cos\theta)+(1)/(sin^2\theta\ cos^2\theta)\n\n=1+(2)/(sin\theta\ cos\theta)+(1)/(sin^2\theta\ cos^2\theta)\n\n=1^2+2sec\theta\ cosec\theta+sec^2\theta\ cosec^2\theta\n\n=(1+sec\theta\ cosec\theta)^2=R$

What is the quotient?1462 ÷ 58

Enter your answer as a mixed number in the simplest form.

Answers

Answer:

25 6/29

Step-by-step explanation:

use long division, and cancel common factors

Answer:

25.20

Step-by-step explanation:

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