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borishaifa [10]
3 years ago
8

Javier volunteers in community events each month. He does not do more than five events in a month. He attends exactly five event

s 25% of the time, four events 30% of the time, three events 20% of the time, two events 15% of the time, one event 5% of the time, and no events 5% of the time. Find the probability that Javier volunteers for less than three events each month. P (x < 3) = 2 Find the expected number of events Javier volunteers in a month. 3.6 It is given that x must be below a certain value, which limits the rows to use in the PDF table. What is the sum of the probabilities of those rows?
Mathematics
1 answer:
enot [183]3 years ago
3 0

Answer:

P(x < 3) = 25\%

E(x) = 3

Step-by-step explanation:

The given parameters can be represented as:

\begin{array}{ccccccc}x & {5} & {4} & {3} & {2} & {1}& {0} & P(x) & {25\%} & {30\%} & {20\%} & {15\%} & {5\%} & {5\%} \ \end{array}

Solving (a): P(x < 3)

This is calculated as:

P(x < 3) = P(x = 0) + P(x = 1) + P(x =2) ----- i.e. all probabilities less than 3

So, we have:

P(x < 3) = 5\% + 5\% + 15\%

P(x < 3) = 25\%

Solving (b): Expected number of events

This is calculated as:

E(x) = \sum x * P(x)

So, we have:

E(x) = 5 * 25\% + 4 * 30\% + 3 * 20\% + 2 * 15\% + 1 * 5\% + 0 * 5\%

E(x) = 125\% + 120\% + 60\% + 30\% + 5\% + 0\%

E(x) = 340\%

Express as decimal

E(x) = 3.40

Approximate to the nearest integer

E(x) = 3

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VikaD [51]

Answer: 8 oz..

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Step-by-step explanation:

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3 years ago
What is the median of 30,33,38,40,51,57,81,91?
enot [183]

Answer:{ \left[\begin{array}{ccc}45.5\\\\\end{array}\right] }{}\\\\\end{array}\right] } ----

Step-by-step explanation: Hey, Basically we are going to arrange the data in an ascending order and the median is the middle value. If the number of values is an even number, the median will be the average of the two middle numbers.

Hope this helps!

4 0
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A salesperson earns a commission of $308 for selling $ 2200 in merchandise. Find the commission rate.
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Step-by-step explanation

5 0
2 years ago
Franco read 3/8 of a chapter of his history book in 1/5 of an hour. At this rate, how many chapters of his history book can he r
Irina18 [472]

He can read 1\frac{7}{8} chapters of his history book in 1 hour.

<em><u>Explanation</u></em>

Franco read 3/8 of a chapter of his history book in 1/5 of an hour.

Suppose, he can read x chapters of his book in 1 hour at this rate.

So, the equation according to the <u>ratio of 'the chapter to the time'</u> will be....

\frac{x}{1}= \frac{\frac{3}{8}}{\frac{1}{5}}\\ \\ \frac{x}{1}=\frac{3}{8}*\frac{5}{1}\\ \\ \frac{x}{1}=\frac{15}{8} \\ \\ x= \frac{15}{8}= 1\frac{7}{8}

So, he can read 1\frac{7}{8} chapters of his history book in 1 hour.

6 0
4 years ago
The accompanying data represent the miles per gallon of a random sample of cars with a​ three-cylinder, 1.0 liter engine.a. Comp
Kobotan [32]

Answer:

1). Z score = 1.464733

Mean = 38.87917

Standard deviation = 3.609405

2). Quartiles:

Q1 = 37.025

Q2 = 38.450

Q3 = 40.800

3). IQR = 3.775

4).

CI (lower fence) = 37.4351

CI (upper fence) =  40.32323

5). There is outlier in the data set. Please see attached box plot for evidence.

Step-by-step explanation:

1). By Z score, we mean:

Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}},

where:

x-bar ==> mean(g) = 38.87917

\mu = 37.80

standard deviation = 3.609405

sample size (n) = 24.

2). By quantile, we mean:

Q = L + (i*(n/4) - Cf)*c

Where L is the lower class limit of the quartile class

Cf is the cumulative frequency before the quartile class

c  is the class size.

3) . IQR = Q3 - Q1

4). CI = \mu \pm *Z_{\alpha/2} \frac{\sigma}{\sqrt{n}},

Where Z_{\alpha/2}  ==> 1.96

In order to replicate and obtain the result, please use the R code below:

g = c(31.5, 36.0, 37.8, 38.5, 40.1, 42.2,34.2, 36.2, 38.1, 38.7, 40.6, 42.5,34.7, 37.3, 38.2, 39.5,  

41.4, 43.4,35.6, 37.6, 38.4, 39.6, 41.7, 49.3)

boxplot(g)

Z = (mean(g) - 37.8)/(sd(g)/sqrt(length(g)))

mean(g)

quantile(g)

IQR(g)

CIl = mean(g) - 1.96*(sd(g)/sqrt(length(g)))

CIU = mean(g) + 1.96*(sd(g)/sqrt(length(g)))

4 0
3 years ago
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