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, "\\mu=60,\\sigma=5"
"X\\sim Bin(\\mu,\\sigma)"
1. "P(X>70)=P(z>\\dfrac{70-60}{5})=P(z>2)"
"=1-P(z\\le2)=1-0.97725=0.02275"
2. "P(61\\le X\\le 65)=P(\\dfrac{61-60}{5}\\le z\\le \\dfrac{65-60}{5})"
"=P(0.2\\le z\\le1)=P(z\\le1)-P(z\\le0.2)=0.84134-0.57926\n\\\\=0.26208"
3. "P(X<55)=P(z<\\dfrac{55-60}{5})=P(z<-1)=0.15866"
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