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2. At a large department store, the average number of years ofemployment for a c

ID: 2917888 • Letter: 2

Question

2. At a large department store, the average number of years ofemployment for a casher is 5.7 with a standard deviation of 1.8years. If an employee is picked at random, what is the probabilitythat the employee has worked at the store for over 10 years. 3. The average hourly wage of workers at a fast foodrestaurant is $5.88/hr with a standard deviation of $0.49. Assumethat the distribution is normally distributed. If a worker at thefast food restaurant is selected at randon, what is the probabilitythat the worker earns more than $4.90. Express your answer inpercent rounded to the nearest integer. 4.Mrs. Smith reading clas can read a mean of 173 word perminute with a standard deviation of 20 words per minute. The top 8%of the class is to receive a special award. What is the miniumnumber of words per minute a student would have to read in order toget the award. Round answer to the nearest integer. 2. At a large department store, the average number of years ofemployment for a casher is 5.7 with a standard deviation of 1.8years. If an employee is picked at random, what is the probabilitythat the employee has worked at the store for over 10 years. 3. The average hourly wage of workers at a fast foodrestaurant is $5.88/hr with a standard deviation of $0.49. Assumethat the distribution is normally distributed. If a worker at thefast food restaurant is selected at randon, what is the probabilitythat the worker earns more than $4.90. Express your answer inpercent rounded to the nearest integer. 4.Mrs. Smith reading clas can read a mean of 173 word perminute with a standard deviation of 20 words per minute. The top 8%of the class is to receive a special award. What is the miniumnumber of words per minute a student would have to read in order toget the award. Round answer to the nearest integer.

Explanation / Answer

2. X~Normal(=5.7, =1.8) P(X>10)=P((X-)/ > (10-5.7)/1.8)=P(Z>2.39 ) =0.0084 3.X~Normal(=5.88, =0.49) P(X>4.9)= P(Z> (4.9-5.88)/0.49) = P(Z>-2)=0.9772 (97%) 4. X~Normal(=173, =20) P(X P(Z< (c-173)/20) = 0.92 -->(c-173)/20 = 1.405 -->c=173+20*1.405=201.1