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Vinil7 [7]
3 years ago
5

Assume the number of customers who order a dessert with their meal on a given night at a local restaurant has the probability di

stribution given below. The variance for the random variable x is _____. a. .77
Mathematics
1 answer:
Bumek [7]3 years ago
5 0

Complete question :

X: 3, 4, 5, 6

f(x): .25, .55, .15, .05

Answer:

0.60

Step-by-step explanation:

Given the data:

Variance (s) :

Σx²*f(x) - E(x)²:

E(x) = (3*0.25) + (4*0.55)+(5*0.15)+(6*0.05) = 4

Σx²*f(x) - E(x)²: [(3^2 * 0.25)+(4^2 * 0.55)+(5^2 * 0.15)+(6^2 * 0.05)] - 4²

16.6 - 16 = 0.6

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I need help asap please!!!
Mashcka [7]
1. x = 118      2. x = -54      3. x = -54      4. x = -54

5 0
3 years ago
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
Mrs. Lane took a survey of the types of pants her students were wearing. She collected the data: jeans 14, shorts 9, and capris
astra-53 [7]
36%
14+9+2=25
9/25=.36
3 0
3 years ago
How many times dose 20 go into 159
inysia [295]
13 times with a remainder of three
8 0
3 years ago
Explain how you can use equivalent fractions to find the quotient of 2 3 ÷ 4.
Virty [35]

Answer:

see below

Step-by-step explanation:

2/3 ÷ 4

We use copy dot flip

The flip means make a reciprocal of the second number

2/3 * 1/4

Multiply the numerators

2*1 = 2

Multiply the denominators

3*4 =12

Put the numerator over the denominator

2/12

Simplify

1/6

8 0
3 years ago
Read 2 more answers
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