A group of statistics students decided to conduct a survey at their university t

Question

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spend studying per week. Based on a simple random sample, they surveyed 141 students. The statistics showed that students studied an average of 25 hours per week with a standard deviation of 12 hours. What is the probability that average student study time is between 23 and 27 hours?
A) 0.9522 B) 0.2440 C) 0.4880 D) 0.0239 A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spend studying per week. Based on a simple random sample, they surveyed 141 students. The statistics showed that students studied an average of 25 hours per week with a standard deviation of 12 hours. What is the probability that average student study time is between 23 and 27 hours?
A) 0.9522 B) 0.2440 C) 0.4880 D) 0.0239 A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spend studying per week. Based on a simple random sample, they surveyed 141 students. The statistics showed that students studied an average of 25 hours per week with a standard deviation of 12 hours. What is the probability that average student study time is between 23 and 27 hours?
A) 0.9522 B) 0.2440 C) 0.4880 D) 0.0239

Explanation / Answer

let X be the random variable denoting the amount of time spent by students studying per week.

assuming X~N((25,122)

we have a sample of size n=141 from this population and Xbar be their average.

so Xbar~N((25,122/141)

now P[23<Xbar<27]=P[(23-25)/(12/sqrt(141))<(X-25)/(12/sqrt(141))<(27-25)/(12/sqrt(141))]

                            =P[-1.979<Z<1.979]

                            =P[Z<1.979]-P[Z<-1.979]=0.976092-0.0239080=0.9522 [option A] [answer]

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