What is J=4rb solved for r?

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.

Ulleksa [173]

Answer:

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

Step-by-step explanation:

We are given the following information in the question:

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.

The distribution function can be written as:

$f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}$

The probability for exponential distribution is given as:

$P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}$

a) P( time between the next two calls will be 54 seconds or less)

$P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763$

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

b) P(time between the next two calls will be greater than 118.5 seconds)

$p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285$

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

6 0

3 years ago

What is the value of x if y = 4

enyata [817]

Answer:

x is infinite

Step-by-step explanation:

x is a function so

if y=f:x

therefore

$0 < x < \infty$

5 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.345

n = sample of participants = 3532

p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

<u>95% confidence interval for p</u>= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0

3 years ago

I need the answers to #1,#2 and #3

Ede4ka [16]

The answers are #1 45 #2 24 #3 10

6 0

3 years ago

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o

blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown. Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860. Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

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Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

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4 0

2 years ago