Question #216689

A sample size of two bulbs of a company was tested and the average of the bulb ways 2000hrs and standard deviation was to be 50hrs. Estimate the number of bulbs of that company are likely to grow
a. More than 2100 hrs
b. Less than 18550 hrs
c. More than 1900 hrs but less than 2050 hrs

Expert's answer

Let X=X= the life time of the bulb: XN(μ,σ).X\sim N(\mu, \sigma).

Given μ=2000 h,σ=50 h.\mu=2000\ h, \sigma=50\ h.

a.

P(X>2100)=1P(X2100)P(X>2100)=1-P(X\leq2100)

=1P(Z2100200050)=1P(Z2)=1-P(Z\leq\dfrac{2100-2000}{50})=1-P(Z\leq2)

0.022750\approx0.022750

0.022750(2500)=570.022750(2500)=57

57 bulbs.


b.

P(X<1850)=P(Z<1850200050)P(X<1850)=P(Z<\dfrac{1850-2000}{50})

=P(Z<3)0.001345=P(Z<-3)\approx0.001345

0.001345(2500)=30.001345(2500)=3

3 bulbs.


c.


P(1900<X<2050)=P(X<2050)P(X1900)P(1900<X<2050)=P(X<2050)-P(X\leq1900)

=P(Z<2050200050)P(Z1900200050)=P(Z<\dfrac{2050-2000}{50})-P(Z\leq\dfrac{1900-2000}{50})

=P(Z<1)P(Z2)=P(Z<1)-P(Z\leq-2)

0.84134470.0227501\approx0.8413447-0.0227501

0.818595\approx0.818595

0.818595(2500)=20460.818595(2500)=2046


2046 bulbs.



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