The weekly salary paid to employees of a small company that supplies part-time laborers averages $750 with a standard deviation of $450. (a) if the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week. (b) if every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week? (c) if every employee receives a 5% salary increase for the next year, how does the normal model change? (d) if the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate? (a) if the weekly salaries are normally distributed, the fraction of employees that make more than $300 per week is approximately nothing. (type an integer or a fraction.)

(a) The fraction of employees is 0.84.

(b)

(c)

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally distributed, estimate the fraction of employees that make more than $300 per week.

We have to calculate the z-value and compute the probability

(b) If every employee receives a year-end bonus that adds $100 to the paycheck in the final week, how does this change the normal model for that week?

The mean of the salaries grows $100.

The standard deviation stays the same ($450)

(c) If every employee receives a 5% salary increase for the next year, how does the normal model change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

The standard deviation increases by a 5% ($472.5)

(d) If the lowest salary is $300 and the median salary is $525, does a normal model appear appropriate?

No. The left part of the distribution would be truncated too much.

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

click to let others know, how is it

0.0

0 votes

1

report

read more on -

step-by-step explanation: