The probability of women is distributed with the mean of 60kg and standard deviation 5kg. What is the probability that chosen at random
1. Greater than 70
2. Between 61&65
3. Less than 55
Solution:
Given, μ=60,σ=5\mu=60,\sigma=5μ=60,σ=5
X∼Bin(μ,σ)X\sim Bin(\mu,\sigma)X∼Bin(μ,σ)
1. P(X>70)=P(z>70−605)=P(z>2)P(X>70)=P(z>\dfrac{70-60}{5})=P(z>2)P(X>70)=P(z>570−60)=P(z>2)
=1−P(z≤2)=1−0.97725=0.02275=1-P(z\le2)=1-0.97725=0.02275=1−P(z≤2)=1−0.97725=0.02275
2. P(61≤X≤65)=P(61−605≤z≤65−605)P(61\le X\le 65)=P(\dfrac{61-60}{5}\le z\le \dfrac{65-60}{5})P(61≤X≤65)=P(561−60≤z≤565−60)
=P(0.2≤z≤1)=P(z≤1)−P(z≤0.2)=0.84134−0.57926=0.26208=P(0.2\le z\le1)=P(z\le1)-P(z\le0.2)=0.84134-0.57926 \\=0.26208=P(0.2≤z≤1)=P(z≤1)−P(z≤0.2)=0.84134−0.57926=0.26208
3. P(X<55)=P(z<55−605)=P(z<−1)=0.15866P(X<55)=P(z<\dfrac{55-60}{5})=P(z<-1)=0.15866P(X<55)=P(z<555−60)=P(z<−1)=0.15866
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment