Question

If 20% of the memory chips made in a certain plant are defective,find the probability that in a lot of 100 randomly chosen chips for inspection
(i)at most 15 chips will be defective,
(ii)the number of defectives will be between 15 and 25.

Answer 1

Answer on Question #86142 – Math – Statistics and Probability

Question

If 20% of the memory chips made in a certain plant are defective, find the probability that in a lot of 100 randomly chosen chips for inspection:

(i) at most 15 chips will be defective

(ii) the number of defectives will be between 15 and 25.

Solution

The binomial distribution $b(x; n, p)$

$$b(x; n, p) = \binom{n}{x} p^x (1 - p)^{n - x}$$

$n = 100, p = 0.2$

(i) at most 15 chips will be defective

$$\begin{array}{l} p(x \leq 15) = \sum_{i=0}^{15} p(x = i) = \binom{100}{0} p^0 (1 - p)^{100 - 0} + \binom{100}{1} p^1 (1 - p)^{100 - 1} + \\ + \binom{100}{2} p^2 (1 - p)^{100 - 2} + \binom{100}{3} p^3 (1 - p)^{100 - 3} + \binom{100}{4} p^4 (1 - p)^{100 - 4} + \\ + \binom{100}{5} p^5 (1 - p)^{100 - 5} + \binom{100}{6} p^6 (1 - p)^{100 - 6} + \binom{100}{7} p^7 (1 - p)^{100 - 7} + \\ + \binom{100}{8} p^8 (1 - p)^{100 - 8} + \binom{100}{9} p^9 (1 - p)^{100 - 9} + \binom{100}{10} p^{10} (1 - p)^{100 - 10} + \\ + \binom{100}{11} p^{11} (1 - p)^{100 - 11} + \binom{100}{12} p^{12} (1 - p)^{100 - 12} + \\ + \binom{100}{13} p^{13} (1 - p)^{100 - 13} + \binom{100}{14} p^{14} (1 - p)^{100 - 14} + \\ + \binom{100}{15} p^{15} (1 - p)^{100 - 15} = 2.027 \times 10^{-10} + 5.096 \times 10^{-9} + 6.302 \times 10^{-8} + \\ + 5.147 \times 10^{-7} + 3.120 \times 10^{-6} + 1.498 \times 10^{-5} + 5.928 \times 10^{-5} + \\ + 1.990 \times 10^{-4} + 5.784 \times 10^{-4} + 0.001478 + 0.003362 + 00.6878 + \\ + 0.012754 + 0.021583 + 0.033531 + 0.048062 \approx 0.1285 \end{array}$$

(ii) the number of defectives will be between 15 and 25.

$$\begin{array}{l} p(15 \leq x \leq 25) = \binom{100}{15} p^{15} (1 - p)^{100 - 15} + \binom{100}{16} p^{16} (1 - p)^{100 - 16} + \\ + \binom{100}{17} p^{17} (1 - p)^{100 - 7} + \binom{100}{18} p^{18} (1 - p)^{100 - 18} + \\ + \binom{100}{19} p^{19} (1 - p)^{100 - 19} + \binom{100}{20} p^{20} (1 - p)^{100 - 20} + \\ + \binom{100}{21} p^{21} (1 - p)^{100 - 21} + \binom{100}{22} p^{22} (1 - p)^{100 - 22} + \\ + \binom{100}{23} p^{23} (1 - p)^{100 - 23} + \binom{100}{24} p^{24} (1 - p)^{100 - 24} + \\ \end{array}$$ $$\begin{array}{l} + \binom{100}{25} p^{25} (1 - p)^{100 - 25} = 0.048062 + 0.063832 + 0.078851 + \\ + 0.090898 + 0.098074 + 0.099300 + 0.094572 + 0.084900 + \\ + 0.090898 + 0.071980 + 0.057734 + 0.043878 \approx 0.8321 \end{array}$$

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