Question #152032
Assuming the lasting time of laptop batteries are normally distributed with a mean of 11 hours and a standard deviation of ) 0.7 hours.
a) If we randomly select 8 batteries, what is the probability that the average lasting time of these laptops is longer than 11.5 hours.
b) In a group of 30 laptops, approximately how many of them will last less than 10 hours?
c) Knowing that a lasting time of a laptop is 20% below the third quartile, how many hours can this laptop last?
1
Expert's answer
2020-12-24T17:41:04-0500

Let X=X= the lasting time of laptop batteries: XN(μ,σ2)X\sim N(\mu, \sigma^2)

Given μ=11h,σ=0.7h\mu=11h, \sigma=0.7h

a) XˉN(μ,σ2/n)\bar{X}\sim N(\mu,\sigma^2/n)

Given n=8n=8


P(Xˉ>11.5)=1P(Xˉ11.5)P(\bar{X}>11.5)=1-P(\bar{X}\leq11.5)

=1P(Z11.5110.7/8)1P(Z2.020305)=1-P(Z\leq\dfrac{11.5-11}{0.7/\sqrt{8}})\approx1-P(Z\leq2.020305)

10.978324=0.021676\approx1-0.978324=0.021676

P(Xˉ>11.5)=0.021676P(\bar{X}>11.5)=0.021676


b)


P(X<10)=P(Z<10110.7)P(X<10)=P(Z<\dfrac{10-11}{0.7})

P(Z<1.428571)0.076564\approx P(Z<-1.428571)\approx0.076564

30(0.076564)=230(0.076564)=2


(c) The third quartile Q3 is the 75th percentile of a data set. 


P(Z<X110.7)=0.75P(Z<\dfrac{X-11}{0.7})=0.75

X110.70.674490\dfrac{X-11}{0.7}\approx0.674490

X11+0.7(0.674490)=11.472143X\approx11+0.7(0.674490)=11.472143

80%100%11.472143=9.18(h)\dfrac{80\%}{100\%}\cdot11.472143=9.18(h)

9.18 hours



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