only draw the graph of the normal curve..
(the life of as army shoes is normally distributed with the mean 8 months and s.d is 2 months. if 5000 pairs are issued how many pairs would be expected to need replacement within 12 months)
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hey mate here is your answer..
Standard normal distribution:
In a standard normal distribution μ=0,σ2=1μ=0,σ2=1
The random variable XX can be converted to the standard normal variable ZZ by the transformation
Z=X−μσ
step by step;-
Step 1:
Let XX be the random variable denoting the life of a pair of army shoes.
X∼N(8,22)X∼N(8,22)
Step 2:
A pair of shoes would require replacement within 12 months if its life is less than 12 months.
∴∴ we have to find P(X<12)
Step 3:
Let ZZ be the standard normal variate.
Z=X−μσZ=X−μσ
=X−82=X−82
When X=12X=12
=12−82=12−82
=42=42
=2=2
Step 4:
∴P(X<12)=P(Z<2)∴P(X<12)=P(Z<2)
=0.5+P(0<Z<2)=0.5+P(0<Z<2)
=0.5+0.4772=0.5+0.4772
=0.9772
Step 5:
Out of 5000 pairs (N=5000) the number of pairs that are expected to be replaced within 12 months=5000×0.9772
⇒4886
hope it helps you dear..
Standard normal distribution:
In a standard normal distribution μ=0,σ2=1μ=0,σ2=1
The random variable XX can be converted to the standard normal variable ZZ by the transformation
Z=X−μσ
step by step;-
Step 1:
Let XX be the random variable denoting the life of a pair of army shoes.
X∼N(8,22)X∼N(8,22)
Step 2:
A pair of shoes would require replacement within 12 months if its life is less than 12 months.
∴∴ we have to find P(X<12)
Step 3:
Let ZZ be the standard normal variate.
Z=X−μσZ=X−μσ
=X−82=X−82
When X=12X=12
=12−82=12−82
=42=42
=2=2
Step 4:
∴P(X<12)=P(Z<2)∴P(X<12)=P(Z<2)
=0.5+P(0<Z<2)=0.5+P(0<Z<2)
=0.5+0.4772=0.5+0.4772
=0.9772
Step 5:
Out of 5000 pairs (N=5000) the number of pairs that are expected to be replaced within 12 months=5000×0.9772
⇒4886
hope it helps you dear..
1RADHIKAA1:
graph??
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