After the first essay examination in a class, a random sample of five papers was selected. The grades were 60, 75, 80, 70, and 90. What would be the 90% confidence interval for the mean grade of all the students?

68.6433≤x≤82.3568

Explanation

confidence interval formula is expressed as

CI = xbar ± z × S/√n

xbar is the mean

z is the z score at 90% = 1.645

s is the Standard deviation

N is the sample size = 5

xbar = 60+75+80+70+ 90/5

xbar = 375/5

xbar = 75

Get standard deviation

SD =√\sum(x-xbar)²/N

SD = √(60-75)²+(75-75)²+(80-75)²+(70-75)²+(90-75)²/5

SD = √15²+0²+5²+5²+15²/5

SD = √225+50+225/5

SD = √500/5

SD =√100

SD = 10

Substitute the derived values into the formula

CI = 75±1.645 × 10/√5

CI = 75 ± (1.645×4.4721)

CI = 75±7.3568

CI = (75-7.3568, 75+7.3568)

CI = (67.6433, 82.3568)

Hence the 90% confidence interval of the student is:

68.6433≤x≤82.3568

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

answer: i wpould go with c

step-by-step explanation: