Assume the cost of an extended 100,000 mile warranty for a particular SUV follow

Question

Assume the cost of an extended 100,000 mile warranty for a particular SUV follows the normal distribution with a mean of $1,380 and a standard deviation of $65. Complete parts (a) through (d) below. a) Determine the interval of warranty costs from various companies that are one standard deviation around the mean. The interval of warranty costs that are one standard deviation around the mean ranges from $11 to $ Type integers or decimals. Use ascending order.) b) Determine the interval of warranty costs from various companies that are two standard deviations around the mean. The interval of warranty costs that are two standard deviations around the mean ranges from $to $ Type integers or decimals. Use ascending order.) c) Determine the interval of warranty costs from various companies that are three standard deviations around the mean. The interval of warranty costs that are three standard deviations around the mean ranges from S to s (Type integers or decimals. Use ascending order.) d) An extended 100,000 mile warranty for this type of vehicle is advertised at $1,640. Based on the previous results, what conclusions can you make? OA. The $1,640 cost of this warranty must be an error in the advertisement because a data value cannot be more than three standard deviations from the mean

Explanation / Answer

a) mean = 1380 , sd= 65

within 1 sd = (1380-65,1380+65)

= (1315,1445)

b) within 2 sd

(1380-2*65,1380+2*65)

=(1380-130,1380+130)

=(1250,1510)

c)

3 sd within mean

(1380-3*65,1380+3*65)

=(1380-195,1380+195)

=(1185,1575)

d) 1640 > 1575

hence it is unusual

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.