Question #133962

5 Gem stones from a certain mine have weights, X grams, which are normally distributed with mean 1.9 g and standard deviation 0.55 g. These gem stones are sorted into three categories for sale depending on their weights, as follows.

Small: under 1.2 g Medium: between 1.2 g and 2.5 g Large: over 2.5 g

(i) Find the proportion of gem stones in each of these three categories.

(ii) Find the value of k such that Pk < X < 2.5 =


1
Expert's answer
2020-09-21T14:51:15-0400

XN(μ,σ2)X\sim N(\mu,\sigma^2)

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

Given μ=1.9g,σ=0.55g\mu=1.9g,\sigma=0.55g


P(X<1.2)=P(Z<1.21.90.55)=P(X<1.2)=P(Z<\dfrac{1.2-1.9}{0.55})==P(Z<1411)0.101577= P(Z<-\dfrac{14}{11})\approx0.101577

P(X<2.5)=P(Z<2.51.90.55)=P(X < 2.5)=P(Z<\dfrac{2.5-1.9}{0.55})=

=P(Z<1211)0.862344= P(Z<\dfrac{12}{11})\approx0.862344

P(1.2<X<2.5)=P(X<2.5)P(X<1.2)P(1.2<X<2.5)=P(X<2.5)-P(X<1.2)\approx

0.8623440.1015770.760767\approx0.862344-0.101577\approx0.760767

P(X2.5)=1P(X<2.5)P(X\geq2.5)=1-P(X<2.5)\approx

10.8623440.137656\approx1-0.862344\approx0.137656

10.1577%, 76.0767%, 13.7656%10.1577\%, \ 76.0767\%,\ 13.7656\%



P(k<X<2.5)=P(X<2.5)P(X<k)=0.8P(k<X<2.5)=P(X<2.5)-P(X<k)=0.8

P(X<k)=P(X<2.5)0.8P(X<k)=P(X<2.5)-0.8\approx

0.8623440.8=0.062344=P(Z<k1.90.55)\approx0.862344-0.8=0.062344=P(Z<\dfrac{k-1.9}{0.55})

z=k1.90.551.53539z=\dfrac{k-1.9}{0.55}\approx-1.53539

k1.90.551.535391.0555k\approx1.9-0.55\cdot1.53539\approx1.0555

k=1.0555k=1.0555


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