Question

QUESTION 3 The human resource officer at a large university constructed the following table. It details the number of sick days in January taken by permanent employees. Number of sick days Percentage of staff 0 69. 9 1 1 4.6 2 7.7 3 2.2 4 2 5 or more 0.2 If there were 699 permanent employees on the payroll, how many took more than one sick day during January?(round to the nearest whole number -please make sure to use conventional rounding rules ie round up if the decimal point in your answer is.5 or greater) QUESTION 4 The mean and standard deviation of a binomial distribution with n=25 and p = 0.20 are C5 and 2 C8 and 4 4 and 3 C5 and 4

Answer 1

Step 1. We have to find the percentage of employees who has taken 4 sick days.

The percentage of staff is 100.

The current available percentage of staff is 94.6(69.9+14.6+7.7+2.2+0.2)

Therefore the percentage of staff which has taken 4 sick days = 100 - 94.6 = 5.4

Step 2. To find out the total number of employees who took more than one sick day

For that, add the percentage of staff who has taken 2 days and more which is equal to 15.5% (7.7+2.2+5.4+0.2)

Now, to find the number of permanent employees = 699× 15% = 108.34.

Rounding up the value we get 108.

Question 4.

Mean of the binomial distribution = n×p = 25× 0.20 = 5

Standard deviation of the binomial distribution = √npq

q = 1- p = 1- 0.20 =0.8

√npq = √25×0.20×0.80 = √4 = 2

The answer is 5 and 2