Suppose that, based on historical data, we believe that the annual percentage salary increases for the chief executive officers of all midsize corporations are normally distributed with a mean of 12.2% and a standard deviation of 3.6%. A random sample of nine observations is obtained from this population, and the sample mean is computed. What is the probability that the sample mean will be greater than 14.4%?

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Helen Rodriguez · Accepted Answer

Step 1  Given Information:  Meanμ=12.2%=0.122  Standard deviationσ=3.6%=0.036  Sample size(n)=9  To find the probability that the sample mean will be greater than 14.4%:  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  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 mean is approximately normal with mean μx―=μ=0.122 and standard deviation of σx―=σn=0.0369=0.012  Step 2  Required probability is obtained as follows:  P(X―&amp;gt;0.144)=P(X―−μx―σx―&amp;gt;0.144−0.1220.012)  =P(Z&amp;gt;1.8333)  =1−P(Z&amp;lt;1.83)   =1−0.96638 [Using sing standard normal table]  =0.03362  Using a standard normal table, look up for z-score of 1.83 i.e., 1.8 in the row and .03 along the column, the intersection of both values gives a probability of 0.96638. Then, subtract the value from total probability 1 (Since, we want the area to the right of z-score and a standard normal table gives area to the left of z-score)  Or using Excel function &quot;=NORM.S.DIST(1.8333,TRUE)&quot;, area corresponding to the left of z-score 1.8333 is 0.966621. Then, subtract the value from 1. That is, 1−0.966621=0.033379  Therefore, probability that the sample mean will be greater than 14.4% is 0.03338

Suppose that, based on historical data, we believe that the annual per

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