Use yout calculator OIS. SnOW your work or explain how you arrived at your concl

Question

Use yout calculator OIS. SnOW your work or explain how you arrived at your conclusion whenever appropriate In a random sample of 175 community college students, the mean number of hours spent studying per week is 13 7 hours and the standard deviation is 4 hours a) Find the standard score (z-score) for students who study the following hours in a particular week. Round Z to the nearest hundredth and interpret the meaning of each answer as it pertains to this problem. ) 22 hours n) 6 hours b)Assuming the distribution of the number of hours y college students study per week is normally distributed, approximately how many of the students in the sample study between 9.7 and 17 7 hours per week? c) Assuming the distribution of the number of hours community college students study per week is many of the students in the sample study less than 5.7 hours per week? d) Without assuming anything about the distribution of the number of hours community college students study per week at least wha percentage (approximately) of the students study between 7.3 and 20.1 hours per week? e) Without assuming anything about the distribution of the number of hours community college students study per week at most what percentage of the students study less than 1.7 hours or more than 25 7 hours per week? ELE 8 9 5 6

Explanation / Answer

a) as we know that z score =(X-mean)/std deviation

hence (i) for 22 hours ; zscore =(22-13.7)/4= 2.075

(ii)for 6 hours ; zscore =(6-13.7)/4=-3.425

b) P(9.7<X<17.7)=P((9.7-13.7)/4<Z<(17.7-13.7)/4)=P(-1<Z<1) =0.8413-0.1587 =0.6827

hence expected number of students =-np =175*0.6827=119.47

c)

P(X<5.7)=P(Z<(5.7-13.7)/4)=P(Z<-2) =0.02275

hence expected number of students =-np =175*0.02275 =3.98

d)asP(7.3<X<20.1)=P(-1.6<Z<1.6

from Chebychev's theorum: P(7.3<X<20.1) >=1-1/k2 =1-1/(1.6)2 =0.6094 ~ 60.94%

e)

asP(1.7<X<25.7)=P(-3<Z<3)

from Chebychev's theorum: P(1.7<X<25.7) >=1-1/k2 =1-1/(3)2 =0.8889 ~ 88.89%

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