Question

In: Statistics and Probability

With R coding Obs: it supposed to use probability density function like X ~ Binomial( n...

With R coding Obs: it supposed to use probability density function like X ~ Binomial( n ,p ) dbinom(X=?, n, prob) pbinom(X=?, n, prob) rbinmo(幾個符合二項分配的X, n, prob) X~Poisson (lamda) dpois(X=?, lamda) ppois (X=?, lamda) rpois (X, lamda)

Exercises 1) A food company produces canned food, and the weight of each can on the production line is in accordance with normal distribution. N (175, 102) (Unit: gram) (a) Specifications indicate that each can weight lless than 155, what is the possibility of randomly choosing a can and it being returned? (b) b) randomly pick a can, What is the probability of qualified? (c) If the company wants the possibility of the product being returned to be <2%, they should set the return policy to allow returns when the van weighs less than how much? (d) (Randomly pick a can, What is the probability of picking a can with a weight between 160 and 190?

2. X~B(n=100,p=0.1) (a) P(125 and n(1-p)>5, X will be approximated by a normal distribution, E(X)=np、Var(X)=np(1-p) calculate P(12100 and np<5, X will be approximated by a poisson distribution, E(X)=np calculate P(12

Solutions

Expert Solution

#1)

Let x be the weight of can, x follows normal distribution with mean µ = 175 and standard deviation σ = 10

Specifications : Weigh of can is must be greater than 155 , otherwise it would be returned.

To find area less than x we use the following function in R:

pnorm(x, mean , sd , lower.tail = TRUE)

To find area greater than x we use the following function in R:

pnorm(x, mean , sd , lower.tail = FALSE)

a) P( can being returned ) = P( x < 155 )

So we use function: pnorm(155, 175 , 10, lower.tail = TRUE)

Answer : 0.0228

b) P( Can being qualified ) = P ( x > 155 )

pnorm(155, 175, 10 , lower.tail = FALSE )

Answer : 0.9772

c) We are asked to find x such that P( X < x ) =0.02

So we can use the function : qnorm(0.02, 175, 10 ,lower.tail = TRUE)

Answer : 154.4625

d) P( 160 < x < 190 ) = P( x < 190 ) - P( x < 160 )

pnorm(190,175,10,lower.tail = TRUE) - pnorm(160,175,10,lower.tail = TRUE)

Answer : 0.8664


Related Solutions

USE R CODING! Pleaseee I need the code With R coding Obs: it supposed to use...
USE R CODING! Pleaseee I need the code With R coding Obs: it supposed to use probability density function like X ~ Binomial( n ,p ) dbinom(X=?, n, prob) pbinom(X=?, n, prob) rbinmo(幾個符合二項分配的X, n, prob) X~Poisson (lamda) dpois(X=?, lamda) ppois (X=?, lamda) rpois (X, lamda) **Suppose the random variable X obeys the binomial allocation B (n=100, p=0.1) (a) Use X as the binomial allocation to calculate P(12≤X≤14) (b) In practice, when np ≥ 5 and n(1-p) ≥ 5, X will...
A probability density function on R is a function f :R -> R satisfying (i) f(x)≥0...
A probability density function on R is a function f :R -> R satisfying (i) f(x)≥0 or all x e R and (ii) \int_(-\infty )^(\infty ) f(x)dx = 1. For which value(s) of k e R is the function f(x)= e^(-x^(2))\root(3)(k^(5)) a probability density function? Explain.
. The joint probability density function of X and Y is given by ?(?, ?) =...
. The joint probability density function of X and Y is given by ?(?, ?) = { ??^2? ?? 0 ≤ ? ≤ 2, 0 ≤ ?, ??? ? + ? ≤ 1 0 ??ℎ?????? (a) Determine the value of c. (b) Find the marginal probability density function of X and Y. (c) Compute ???(?, ?). (d) Compute ???(?^2 + ?). (e) Determine if X and Y are independent
The random variables X and Y have the following probability density function (pdf). Conditional probability density...
The random variables X and Y have the following probability density function (pdf). Conditional probability density function is f(xly) = ax/y^2 , 0<x<y, 0<y<1 =0 , o/w marginal density finction is f(y) = by^4 ,0<y<1 = 0, o/w At this time, find the constant a and b values ​​to be the probability density function, and then indicate whether the random variables X and Y are independent, and if not, indicate whether they are positive or inverse.
If x is a binomial random​ variable, use the binomial probability table to find the probabilities...
If x is a binomial random​ variable, use the binomial probability table to find the probabilities below. a. P(x<6) for n = 15, p=0.2 b. P(x>=14) for n=20, p=0.8 c. P(x=23) for n=25, p=0.1
If x is a binomial random​ variable, use the binomial probability table to find the probabilities...
If x is a binomial random​ variable, use the binomial probability table to find the probabilities below. a. P(x=3) for n=10, p=0.5 b. P(x≤4) for n=15, p=0.3 c. P(x>1) for n=5, p=0.2 d. P(x<6) for n=15, p=0.8 e. P(x≥14) for n=25, p=0.8 f. P(x=3) for n=20, p=0.1
For a binomial probability distribution, n = 130 and p = 0.60. Let x be the...
For a binomial probability distribution, n = 130 and p = 0.60. Let x be the number of successes in 130 trials. a. Find the mean and standard deviation of this binomial distribution. a. Find the mean and the standard deviation of this binomial distribution. b. Find to 4 decimal places P(x ≤ 75) using the normal approximation. P(x ≤ 75) = c. Find to 4 decimal places P(67 ≤ x ≤ 72) using the normal approximation. P(67 ≤ x...
Use R for coding Fit a density estimate to the data set `pi2000` (**UsingR**). Compare with...
Use R for coding Fit a density estimate to the data set `pi2000` (**UsingR**). Compare with the appropriate histogram. Why might you want to add an argument like `breaks = 0:10-.5` to `hist()`? I know this much code: install.packages("UsingR") library(UsingR) data(pi2000) if you put this in R a data set will appear
Consider a continuous random variable X with the probability density function f X ( x )...
Consider a continuous random variable X with the probability density function f X ( x ) = x/C , 3 ≤ x ≤ 7, zero elsewhere. Consider Y = g( X ) = 100/(x^2+1). Use cdf approach to find the cdf of Y, FY(y). Hint: F Y ( y ) = P( Y <y ) = P( g( X ) <y ) =
The probability density function of the continuous random variable X is given by fX (x) =...
The probability density function of the continuous random variable X is given by fX (x) = kx, (0 <= x <2) = k (4-x), (2 <= x <4) = 0, (otherwise) 1) Find the value of k 2)Find the mean of m 3)Find the Dispersion σ² 4)Find the value of Cumulative distribution function FX(x)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT