Question

show calculator command for 2 please

2. Suppose a random variable, x, arises from a binomial experiment. If n = 22, and p = 0.85, find the following probabilities using the binomial formula. Show calculator command used. a. (2 pts) P(x = 18) = b. (2 pts) P(x 3) = c. (2 pts) P(x220) = 3. The proportion of red M&M's in a milk chocolate packet is approximately 21% (Madison, 2013). Suppose a package of M&M's typically contains 52 M&M's. a. (2 pts) State the random variable. b. (4 pts) Argue that this is a binomial experiment

Answer 1

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(5. 5 4१8 ४10-) + (२. ० १८ ० ४ !0 - 19) (०.० 0. 0 ০ ০ १२२7 3 0. 00৫০ ০ 4৪।8 + /0-13 २.0 १ 6८ ) + 0. 0 5 5 4१४ + २. 5 3 0 x. 10-13 (a) P[x>, २] ८ = P [ x:२७] + P[x৭। ] 4 P [x २२ ] + (३१) ०-15)" 0- 15)' - ( 3१) (o१६ १°(a14 ) २। (१ )(०.15) २२ २२ (७-15) ৫ २२ ) (০. ४5) ২ ৫-1६)° 0. ২ ০145 + 0.10 + 2 + 7 0. 0 २ ४0 0 ও. 33 81+ (own )

2.

We can also do this using technology.

At first go to distribution menu on the TI 83 calculator. There we have to use the command binomcdf(n,p,c) to get the probability that X takes value equal to or less than c (c is a constant, integer), and binompdf(n,p,c) to get the value of P[X=c], where X~ Bin(n,p)

So here,

1. P(X=18) = binompdf(22,0.85,18)

P(X<=3) = binomcdf(22,0.85,3)

c)P(X>=20) = 1 – P(X<20)

                   = 1 – P(X<=19)

                   = 1 – binomcdf(22,0.85,19)

Using command we can find binomial probabilities in this way described above .

3.

Proportion of red M&M’ in a milk chocolate packet is approximately 21%. Here we are supposing that a package of M&M’s typically contains 52 M&M’s.

a.

Here the random variable is number of red M&M’s in a milk chocolate packet as number of red M&M is a variable and a probability is associated with it such that out of 100 M&M’s, we can get 21 red M&M’s.

b.

Proportion of red M&M’s in a milk chocolate packet being 21% , we can consider this proportion as the probability of red M&M in a milk chocolate packet.

Again, there are typically 52 M&M’s in a packet.

If we consider getting red M&M as success and total number of M&M as total number of trials, then here total number of trials is 52 and probability of success is 0.21

Clearly here the random variable X has properties of a binomial random variable because probability of red M&M’s being picked up is 0.21 and we have 52 M&M’s in a packet.

So, the random variable is a binomial random variable with n=52 and p=0.21

Hence, this is a binomial experiment.