select 1) σ or (2) σ/n√ for the following problems A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 94 months with a standard deviation of 9 months. If the claim is true, what is the probability that the mean monitor life would be greater than 93.5 months in a sample of 111 monitors? Suppose cattle in a large herd have a mean weight of 1217 lbs and a standard deviation of 102 lbs. What is the probability that the mean weight of the sample of cows would less than 1150 lbs if 116 cows are sampled at random from the herd? A soft drink machine outputs a mean of 25 ounces per cup. The machine's output is normally distributed with a standard deviation of 4 ounces. What is the probability of filling a cup between 18 and 33 ounces?

The mean is 94 and standard deviation is 9. Let X be the monitor life's.Since the distribution is not know. The sample size is 111 which is large, also the probability is calculated for mean indicating that normal distribution is appropriate with mean μ and standard deviation σn. Correct Answer: Option (2). The probability that the mean monitor life would be greater than 93.5 months in a sample of 111 monitors is, Px¯>93.5=1-Px¯<93.5=1-Px¯-μσn<93.5-949111=1-Pz<-0.5853 The probability of z less than –0.5853 can be obtained using the excel formula “=NORM.S.DIST (–0.5853, TRUE)”. The probability value is 0.2792. The required probability is, Px¯>93.5=1-Pz<-0.5853=1-0.2792=0.7208 Thus, the probability that the mean monitor life would be greater than 93.5 months in a sample of 111 monitors is 0.7208.The mean weight of 1217 lbs and a standard deviation of 102 lbs. Let X be the weight of the cow. Since the distribution is not know. The sample size is 116 which is large, also the probability is calculated for mean indicating that normal distribution is appropriate with mean μ and standard deviation σn. Correct Answer: Option (2). The probability that the mean weight of the sample of cows would less than 1150 lbs if 116 cows are sampled at random from the herd is, Px¯<1150=Px¯-μσn<1150-1217102116=Pz<-7.0746 The probability of z less than –7.0746 can be obtained using the excel formula “=NORM.S.DIST (–7.0746, TRUE)”. The probability value is 0. The required probability is, Px¯<1150=Pz<-7.0746=0 Thus, the probability that the mean weight of the sample of cows would less than 1150 lbs if 116 cows are sampled at random from the herd is 0.Let X be the machine output which follows normal distribution with mean 25 and standard deviation 4 ounces. Since the distribution is know. The probability also calculated by using the mean μ and standard deviation σ because the probability is calculated for only single cup. Correct Answer: Option (1). The probability of filling a cup between 18 and 33 ounces is, P18<X<33=P18-254<X-μσ<33-254=P-1.75<z<2=Pz<2-Pz<-1.75 The probability of z less than –1.75 can be obtained using the excel formula “=NORM.S.DIST (–1.75, TRUE)”. The probability value is 0.0401. The probability of z less than 2 can be obtained using the excel formula “=NORM.S.DIST (2, TRUE)”. The probability value is 0.9772. The required probability is, P18<X<33=Pz<2-Pz<-1.75=0.9772-0.0401=0.9371 Thus, the probability of filling a cup between 18 and 33 ounces is 0.9371.