Answer to Question #127977 in Statistics and Probability for ada

Question

The thickness of books in a bookstore follows a normal distribution with the mean of 45mm and the standard deviation of 12mm.

a) A book is obtained from the bookstore, find the probability that

i) The book is thinner than 30mm.

ii) The book’s thickness is between 34mm and 50mm.

b) 10 books are chosen from the bookstore at random, the mean thickness of that 10 books, 𝑋̅10, is measured. Find the probability that 𝑋̅10 lies between 43mm to 47mm.

c) 𝑛 books are chosen from the bookstore at random, the mean thickness of that 𝑛 books, 𝑋̅𝑛, is measured. Find the smallest value of 𝑛 so that P(44<𝑋̅𝑛<46)≥0.99 .

Answer 1

a)

Let $X=$ the thickness of book: $X\sim N(\mu, \sigma^2)$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

(i)

$\mu=45, \sigma=12$

$$P(X<30)=P(Z<\dfrac{30-45}{12})=P(Z<-1.25)\approx$$

$$\approx0.105650$$

(ii)

$\mu=45, \sigma=12$

$$P(34<X<50)=P(X<50)-P(X\leq34)=$$

$$=P(Z<\dfrac{50-45}{12})-P(Z\leq\dfrac{34-45}{12})\approx$$

$$\approx P(Z<0.416667)-P(Z<-0.916667)\approx$$

$$\approx0.6615389-0.1796587\approx0.481880$$

b)

$\mu=45, n=10,\sigma={12\over \sqrt{10}}$

$$P(43<\bar{X}_{10}<47)=P(\bar{X}_{10}<47)-P(\bar{X}_{10}\leq43)=$$

$$=P(Z<\dfrac{47-45}{{12\over \sqrt{10}}})-P(Z\leq\dfrac{43-45}{{12\over \sqrt{10}}})\approx$$

$$\approx P(Z<0.527046)-P(Z\leq-0.527046)\approx$$

$$\approx0.7009193-(1-0.7009193)\approx0.401839$$

c)

$\mu=45, \sigma={12\over \sqrt{n}}$

Given $P(44<\bar{X}_{n}<46)\geq0.99$

$$P(44<\bar{X}_{n}<46)=P(\bar{X}_{n}<46)-P(\bar{X}_{n}\leq44)=$$

$$=P(Z<\dfrac{46-45}{{12\over \sqrt{n}}})-P(Z\leq\dfrac{44-45}{{12\over \sqrt{n}}})=$$

$$=2\cdot P(Z<\dfrac{\sqrt{n}}{{12}})-1\geq0.99$$

$$P(Z<\dfrac{\sqrt{n}}{{12}})\geq0.995$$

$$\dfrac{\sqrt{n}}{{12}}\geq2.575829$$

$$n\geq956$$

$n=956$