Question

5. -9.09 points DevoreStat8 3.E.069 My Notes Ask Your Each of 12 efrigerators of a certain type has been returned to distributor because of an audible, high-pitched, oscillating noise when the efrigerators are running. Suppose that 7 of these efrigerators ha defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor (a) Calculate P(X4) and PX S 4). (Round your answers to four decimal places.) P(X S 4)0.879 (b) Determine the probability that X exceeds its mean value by more than 1 standard deviation. (Round your answer to four decimal places.) (c) Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X 3) than to use the hypergeometric pmf We can approximate the hypergeometric distribution with the -Selectdistribution if the population size and the number of successes are large. Here n Approximate PX 3) using that method. (Round your answer to three decimal places.) and p MIN You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Read It Talk to aTuter Submit Answer Save Progress Practice Another Version

Answer 1

(a)

Here X has hypergeometric distribution with parameters as follows:

Population size: N=12

Sample size: n =6

Number of defective in population : k = 7

The required probabilites are:

$P(X=4)=\frac{\binom{7}{4}\binom{5}{2}}{\binom{12}{6}}=0.3788$

$P(X\leq 4)=\sum_{x=2}^{4}\frac{\binom{7}{x}\binom{5}{6-x}}{\binom{12}{6}}=0.8788$

(b)

$\mu=\frac{nk}{N}=\frac{6\cdot 7}{12}=3.5$

$\sigma=\sqrt{\frac{nk(N-k)(N-n)}{N^{2}(N-1)}}=\sqrt{\frac{6\cdot 7\cdot 5\cdot 6}{12^{2}\cdot 11}}=0.8919$

The interval corrsponding to $\mu+\sigma =3.5+0.89=4.39$

So requried probability is

$P(X>4.39)=P(X\geq 5)=0.1212$

(c)

Binomial distribution

n=25 and p = 40/400 = 0.1

So

$P(X\leq 3)=\sum_{x=0}^{3}\binom{25}{x}(0.1)^{x}(0.9)^{25-x}=0.7636$