Consider two x distributions corresponding to the same x distribution with u o = 9.5. The first x is based on the samples of size n = based on the samples of n = 95. You calculate the probability P(58 < X < 61) for both x distributions. What can you say about the calculated probabilities of the first and second x = 60 and 40 and the second distribution is distributions? Select one: a. The calculated probability of the second x distribution is higher than the calculated probability of the first x distribution because of the larger sample size taken from the sece x distribution and the standard deviation of the second x distribution is smaller. Oh W e cannot oglouloto

Given Information: Mean μ=60 Standard deviation σ=9.5 Sample size of first x¯ is n = 40 Sample size of second x¯ is n = 95 Z-score is a measure which tells how many standard deviations away, a data value is, from the mean. It is given by the formula: Z=X-μσ Based on the concept of central theorem, if a sample of size n is drawn from a normal population with mean μ and standard deviation σ, then the sampling distribution of sample means is approximately normally distributed with mean μx¯=μ and standard deviation σx¯=σn First sample: Sampling distribution of sample mean is μx¯=μ=60 and standard deviation is σx¯=9.540=1.502081889≈1.5021 Required probability is obtained as follows: P58<X¯<61=PX¯<61-PX¯<58=PX¯-μx¯σx¯<61-601.5021-PX¯-μx¯σx¯<58-601.5021=PZ<0.67-PZ<-1.33=0.74857-0.09176 [using standard normal table]=0.65681 Using standard normal table, look up for z-score 0.67 i.e., 0.6 in the row and .07 along the column, the intersection of both values gives a probability of 0.74857. Similarly, look up for z-score -1.33 i.e., -1.3 in the row and .03 along the column, the intersection of both values gives a probability of 0.09176. Then, subtract the lower probability value from the higher value. (Since, we want the area or probability between two z-scores) Thus, for n = 40, probability that the sample mean lies between 58 and 61 is 0.65681 Second sample: Sampling distribution of sample mean is μx¯=μ=60 and standard deviation is σx¯=9.595=0.9746794345≈0.9747 Required probability is obtained as follows: P58<X¯<61=PX¯<61-PX¯<58=PX¯-μx¯σx¯<61-600.9747-PX¯-μx¯σx¯<58-600.9747=PZ<1.03-PZ<-2.05=0.84849-0.02018 [using standard normal table]=0.82831 Using standard normal table, look up for z-score 1.03 i.e., 1.0 in the row and .03 along the column, the intersection of both values gives a probability of 0.84849 Similarly, look up for z-score -2.05 i.e., -2.0 in the row and .05 along the column, the intersection of both values gives a probability of 0.02018 Then, subtract the lower probability value from the higher value. (Since, we want the area or probability between two z-scores) Thus, for n = 95, probability that the sample mean lies between 58 and 61 is 0.82831 The calculated probability of second x¯ distribution is higher than the calculated probability of the first x¯ distribution because of the larger sample size taken from the second x¯ distribution and the standard deviation of the second x¯distribution is smaller.