A student Surveyed 200 people in a statistics lecture to find out how many hours

Question

A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15% B.0.3% C. 1.5 % D. 2.5 % A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15% B.0.3% C. 1.5 % D. 2.5 % A student Surveyed 200 people in a statistics lecture to find out how many hours per night the class members typically spend sleeping. The data showed a normal distribution with a mean of 6.8 hours and a standard deviation of 0.75 hours.
Which percentage of students reported typically getting less than 4.55 hours of sleep per night ?
A. 0.15% B.0.3% C. 1.5 % D. 2.5 %

Explanation / Answer

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    4.55      
u = mean =    6.8      
          
s = standard deviation =    0.75      
          
Thus,          
          
z = (x - u) / s =    -3      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -3   ) =    0.0015 or 0.15% [OPTION A]

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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