The mean and standard deviation of marks of 500 students in an examination are 52 and 8 respectively. If the marks are normally distributed. Find Number of students getting marks between 48 and About the author Natalia
Answer: Step-by-step explanation: ok so here, n=500 X1=45 and X2=60 mean m=55 std deviation s=12 for normal distribution, we need to convert X1 and X2 to Z1 and Z2 Z1=(X1-m)/s Z2=(X2-m)/s on substituting values we get Z1=-0.83 and Z2=1.25 Now we see values of areas A1(-0.83) and A2(1.25), corresponding to Z1 and Z2 from cumulative normal distribution table. A1=1-0.7967=0.2023 (Since Z1 is Negative) A2=0.8944 Reply
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
ok so here, n=500
X1=45 and X2=60
mean m=55
std deviation s=12
for normal distribution, we need to convert X1 and X2 to Z1 and Z2
Z1=(X1-m)/s Z2=(X2-m)/s
on substituting values we get Z1=-0.83 and Z2=1.25
Now we see values of areas A1(-0.83) and A2(1.25), corresponding to Z1 and Z2 from cumulative normal distribution table.
A1=1-0.7967=0.2023 (Since Z1 is Negative)
A2=0.8944