Suppose that there are twenty employees in an office, and each employee would li
ID: 3302792 • Letter: S
Question
Suppose that there are twenty employees in an office, and each employee would like a new computer with probability 0.4 Each employee is independent of the others. (a) What is the probability that at least four employees would like new computers? (b) What is the probability that at least sixteen employees would like new computers? (c) What is the probability that eight to ten employees would like new computers? d) Suppose that the office manager would like to purchase enough computers such that it is very likely that all employees wanting new computers will receive them, while also keeping costs manageable. To help the manager explore their options, find the smallest integer k such that P(X> k)30.01 when X is a random variable counting the number of employees who want new computers (i.e., k is the smallest number of computers that has at least a 99% likelihood of providing new computers to all employees wanting them).Explanation / Answer
a. There are two possible outcomes-success (employee would like a new computer) and failure (employee would not like a new computer); there are n=24 independnet trials, and probability of success is constant throughout the trials. Therefore, this accounts for binomial distribution. Use, P(X,r)=nCr(p)^r(1-p)^n-r, where, r denotes specific number of success in n trials.
P(X>=4)=1-P(X<4)=1-P(X<=3) [look into binomial table with p=0.4, n=20 and x=3, the table gives values binomial cumulative distribution values, that is P(X<=x)]
b. P(X>=16)=1-P(X<16)=1-P(X<=15)=0.9997
c. P(8<X<10)=P(X<10)-P(X<8)=P(X<=9)-P(X<=7)=0.7553-0.4159=0.3394
d. Looking into binomial table for p=0.4, one can see that for n=8 and x=6, the probability value is 0.01.
That is P(X>k)=1-P(X<=k)=1-P(8,6)=1-0.99=0.01.
Thus, k=6.
=0.0160
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