Question

The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa. If five alloys are randomly selected from a large population of such alloys, what is the probability that what is the probability that two or more of themhavea greater strength than 11GPa

Answer 1

Answer: 0.3461

Explanation:

Given : The strength of an aluminum alloy is normally distributed .

Mean :
`\mu=10\text{ GPa}`

Standard deviation :
`\sigma=1.4\text{ GPa}`

Sample size : n=5

Let x be the random variable that represents the strength of aluminum alloy .

z-score :
`z=(x-\mu)/(\sigma)`

For x= 11 GPa

`z=(11-10)/(1.4)\approx0.71`

Using the standard normal distribution table , we have

The probability that a aluminum alloy has greater strength than 11GPa:

=
`P(x>11)=P(z>0.71)=1-P(\leq0.71)=1-0.7611479=0.2388521\approx0.24`

Now by using the binomial distribution, the probability that two or more of them have a greater strength than 11GPa :-

`P(X\geq2)=1-P(x<2)=1-(P(0)+P(1))\\\\=1-(^5C_0(0.24)^0(1-0.24)^5+^5C_1(0.24)^1(1-0.24)^4)\\\\=1-((0.76)^5+5(0.24)(0.76)^4)=0.3461013504\approx0.3461`

Hence, the required probability = 0.3461