The print on the package of 100-watt General Electric soft-white light-bulbs say

Question

The print on the package of 100-watt General Electric soft-white light-bulbs says that these bulbs have an average life of 750 hours. Assume that the lives of all such bulbs have a normal distribution with a mean of 750 hours and a standard deviation of 55 hours. Find the probability that the mean life of a random sample of 25 such bulbs will be a. greater than 740 hours Round your answer to four decimal places. b. between 725 and 740 hours Round your answer to four decimal places. c. within 15 hours of the population mean Round your answer to four decimal places. d. less than the population mean by at least 5 hours or more Round your answer to four decimal places.

Explanation / Answer

Mean life of the bulbs are = 750 Hours

Standard Deviation of bulb life = 55 Hours

Sample sie = 25

(a) Here we have to calculate the probabiity that mean ife of the sample of 25 bulbs is greater than 740 Hours.

Standard error of the sample mean = (/n)  = 55/ 25 = 11

Pr( Xbar > = 740 ; 750 ; 11) = 1 - Pr( Xbar < 740 ; 750 ; 11)

Here Z - value = (740 - 750)/ 11 = -0.91

so Pr( Xbar > = 740 ; 750 ; 11) = 1 - (-0.91)

where is the cumulative normal probability function

Pr( Xbar > = 740 ; 750 ; 11) = 1 - 0.1814 = 0.8186

(b) Pr( 725 <= Xbar < = 740 ; 750 ; 11) = Pr(Xbar < = 740 ; 750 ; 11) - Pr( Xbar <= 725; 750 ; 11)

Z - values for the given two values are

Z- value for 725= (725 - 750)/11 = - 2.273 and for 740 =  (740 - 750)/ 11 = -0.91

Pr( 725 <= Xbar < = 740 ; 750 ; 11) = (-0.91) - (-2.27) = 0.1814 - 0.0116 = 0.1698

(c) within 15 hours around population mean

so Z - value = (757.5 - 750)/ 11 = 0.68

so Pr( 742.5 <= Xbar < = 757.5 ; 750 ; 11) = (0.68) - (-0.68) = 0.7517 - 0.2483 = 0.5034

(d) less than the population mean by at least 5 hours or more

Pr(X <= 745 ; 750; 11) = ?

Z - value = (745 - 750)/11 = -0.455

Pr(X <= 745 ; 750; 11) = = (-0.455) = 0.3246

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