Some examples from this set of practice questions
1.
You are given that a certain brand of audio speakers has an average trouble-free lifespan of 7.5 years with a standard deviation of 30 months. Assume that the lifetimes of individual speakers in the sample are normally distributed.
Answer: Converting Standard deviation into years [(30 months)/(12 months)]= 2.5 years z= (x-µ )/σ =(9 years-7.5 years)/(2.5 years) =0.6 Using a normal distribution table, we find that the probability corresponding to a z-score of 0.6 is approximately 0.2257. P(X > 9) = P(0.6 > z) = 0.5 – 0.2257 = 0.2743 or 27.43%
2.
Twelve audio speakers of this brand are purchased by a company for 12 of its employees. What is the probability that exactly 7 out of the 12 audio speakers purchased by the company will work for over 9 years without any problem?
Answer: n=12, p=7/12=0.58 , q=1 – 0.58 = 0.42 P(r > 9) = P(10) + P(11) + P(12) =12!/10!(12-10)!(〖0.58)〗^10 (〖0.42)〗^2+ 12!/11!(12-11)!(〖0.58)〗^11 (〖0.42)〗^1+12!/10!(12-12)!(〖0.58)〗^12 (〖0.42)〗^0 =0.0501 + 0.0125 + 0.0014 =0.064 or 6.4%
3.
The annual family expenditure of 421 families in the main city is normally distributed with amean of Rs. 800,000 and a standard deviation of Rs.75, 000. If a family of the city is selectedrandomly,(i) What is the probability that the annual expenditure of a family is greater than Rs600,000?
Answer: P(X>600000)=P((X−800000) / 75000>(600000−800000) / 75000) =P(z>−2.6667) =0.9962
4.
The annual family expenditure of 421 families in the main city is normally distributed with amean of Rs. 800,000 and a standard deviation of Rs.75, 000. If a family of the city is selectedrandomly. (ii) What is the probability that the annual expenditure of a family is in between Rs.650,000 and Rs.840, 000.
Answer: P(650000<X<840000)=P((6500000−800000) / 75000<(X−800000) /75000<(8400000−800000) / 75000) =P(−2<z <0.5333) =P(z<0.5333)−P(z<−2) =0.7031−0.0228 =0.6803
5.
The annual family expenditure of 421 families in the main city is normally distributed with amean of Rs. 800,000 and a standard deviation of Rs.75, 000. If a family of the city is selectedrandomly. (iii) What is the probability that the annual expenditure of a family is in between Rs. 880000 and Rs. 960, 000?
Answer: P(8800000<X<960000)=P((8800000−800000) / 75000<(X−800000) / 75000<(9600000−800000) / 75000)=P(1.0667<z<2.1333) =P(z<2.1333)−P(z<1.0667)=0.9836−0.8570=0.1266
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