Is the five-number summary or the mean and standard deviation a better brief sum

Question

Is the five-number summary or the mean and standard deviation a better brief summary for this distribution? Explain your choice. Which of the following likely to have a mean that is smaller than the median? the salaries of all National Football League players the scores of students out of 100 points on a very easy exam in which most score perfectly, but a few do the prices of homes in the city of Detroit the scores of students out of 100 points on a very difficult in which most score poorly, but a few do After the physical training required during World War II, the distribution of mile runt times for male students at the University of was approximately Normal with mean 7.11 minutes and standard deviation 0.74 minutes. Use the to estimate the answers each of the following question. What proportion of these students could run a mile in 9.33 minutes of less? What proportion of these students could run a mile between 5.63 minutes and 7.85 minutes? Mechanical measurements on supposedly identical objects usually vary. The variation often follows a Normal distribution a Normal distribution. The stress required to break a type of bolt varies normally with mean 75 kilopounds per square inch (ksi) and standard deviation 8.3 ksi. Determine the percent 10 4 decimal places of these bolts that will with standard a stress between 65 and 80 ksi without breaking

Explanation / Answer

Q.6 No distribution given, leaving it

Q. 7 Option B is correct. Lets see each option in this question

In this question it is asked that in which case mean will be smaller than median.

(a) salaraies of NFL players- it can be or it can't be

(c) the prices of homes in detroit - it can be or can't be

(d) the scores of students in a very difficult exam in which most score poor but few did very good, let see there are 10 students, out of 10, 8 scored 10 and rest two scored 90

so mean = 26 and median = 10 so median is smalled then mean

(b) So, the option b is correct: the scores of students in a very easy exam in which most score good but few did very bad, let see there are 10 students, out of 10, 8 scored 90 and two scored 10

so mean = 74 and median = 90

Q.8 Mean = 7.11 mins and standerd deviation = 0.74 mins

P ( x< 9.33 mins) = ?

Z = (9.33- 7.11) / 0.74 = 3

so by 68-95-99.7 rule

P(x < 9.33 minutes) = 0.5 + 0.997/2 = 0.9985

Q. 9 P ( 5.63 < X < 7.85) = ?

for x = 5.63, Z value = (5.63 - 7.11)/0.74 = -2

and for x = 7.85, Z value = ( 7.85 - 7.11) / 0.74 = 1

so P( 5.63 < X < 7.85) by by 68-95-99.7 rule = 95/2 + 68/2 = 0.815 = 81.5 %

Q. 10 Mean = 75 ksi and standerd deviation = 8.3 ksi

so it required to caluclate P (65 <X < 80 ; 75; 8.3) = ?

so for X = 65; Z value = (65 -75) / 8.3 = -1.205

and for X = 80; Z - value = ( 80 - 75)/ 8.3 = 0.602

so P(65 <X < 80 ; 75; 8.3) will be calculated by using z table for these 2 values

= 0.7265 - 0.1141 = 0.6124 = 61.24%

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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