The random variable \( X\sim B(8,0.3) \). Find \( P(X=2) \).
\( P(X=2)=\dbinom{8}{2}(0.3)^2(0.7)^6=28\times0.09\times0.117649\approx0.296 \).
Statistical Distributions
10 exam-style questions on the binomial and normal distributions. Tap Show answer for full working.
3 Easy4 Medium3 Hard
If \( X\sim B(10,0.4) \), find the mean and variance of \( X \).
\( \mathrm{E}(X)=np=4 \). \( \mathrm{Var}(X)=np(1-p)=10\times0.4\times0.6=2.4 \).
\( Y\sim N(50,16) \). Find \( P(Y<46) \) using the standard normal table (or calculator), giving the answer to 3 d.p.
\( Z=\dfrac{46-50}{4}=-1 \). \( P(Z<-1)=1-\Phi(1)\approx1-0.8413=0.159 \).
\( X\sim B(12,0.25) \). Find \( P(X\le2) \).
\( P(X=0)+P(X=1)+P(X=2) \). \( (0.75)^{12}+\dbinom{12}{1}(0.25)(0.75)^{11}+\dbinom{12}{2}(0.25)^2(0.75)^{10}\approx0.0317+0.1267+0.2323\approx0.391 \).
\( X\sim N(\mu,25) \) and \( P(X<40)=0.1587 \). Find \( \mu \).
\( \Phi\left(\dfrac{40-\mu}{5}\right)=0.1587 \) ⇒ \( \dfrac{40-\mu}{5}=-1 \) ⇒ \( 40-\mu=-5 \) ⇒ \( \mu=45 \).
A manufacturer claims that 90% of its light bulbs last more than 1000 hours. A random sample of 15 bulbs is tested. Using a binomial model, find the probability that fewer than 12 last more than 1000 hours.
\( X\sim B(15,0.9) \). \( P(X\le11)=1-P(X\ge12) \). Alternatively compute \( \sum_{k=0}^{11}\dbinom{15}{k}(0.9)^k(0.1)^{15-k} \). Approx. 0.055 (exact calculation recommended with calculator).
\( X\sim N(100,36) \). Find \( a \) such that \( P(X>a)=0.05 \).
\( \Phi\left(\dfrac{a-100}{6}\right)=0.95 \) ⇒ \( \dfrac{a-100}{6}\approx1.645 \) ⇒ \( a\approx100+9.87=109.87 \).
\( X\sim B(20,0.4) \). Use a normal approximation (with continuity correction) to estimate \( P(X\ge10) \).
\( \mathrm{E}(X)=8 \), \( \mathrm{Var}=4.8 \), \( \sigma\approx2.19 \). \( P(X\ge10)\approx P(Y>9.5) \) where \( Y\sim N(8,4.8) \). \( Z=\dfrac{9.5-8}{\sqrt{4.8}}\approx0.685 \). \( P(Z>0.685)\approx0.247 \).
The lengths of metal rods are normally distributed with mean 50 cm and standard deviation 1.2 cm. Rods shorter than 48 cm or longer than 52 cm are rejected. Find the proportion rejected.
\( P(X<48)+P(X>52)=2P\left(Z<\dfrac{48-50}{1.2}\right)=2P(Z<-1.667)\approx2\times0.0478=0.0956 \) (about 9.6%).
\( X\sim N(\mu,\sigma^2) \) with \( P(X<20)=0.05 \) and \( P(X>30)=0.10 \). Find \( \mu \) and \( \sigma \).
\( \dfrac{20-\mu}{\sigma}=-1.645 \), \( \dfrac{30-\mu}{\sigma}=1.282 \). Subtract: \( \dfrac{10}{\sigma}=2.927 \) ⇒ \( \sigma\approx3.416 \). Then \( 20-\mu=-1.645\times3.416 \) ⇒ \( \mu\approx25.62 \).