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Q.In a normal distribution, 31% of the articles are under 45 and 8% are over 64. Calculate the mean and standard deviation of the distribution

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★est
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Turn each percentage into a z-value from the normal table, write the two equations 45−μσ=−0.5\frac{45-\mu}{\sigma}=-0.5 and 64−μσ=1.4\frac{64-\mu}{\sigma}=1.4, and solve them simultaneously to get mean =50=50 and standard deviation =10=10.

Standard normal variable: z=x−μσ\displaystyle z=\frac{x-\mu}{\sigma}

where xx = the given mark, μ\mu = mean (unknown), σ\sigma = standard deviation (unknown). Table values used: for a left-tail area 0.310.31, the between-mean area is 0.5−0.31=0.19⇒z=0.50.5-0.31=0.19\Rightarrow z=0.5; for a right-tail area 0.080.08, the between-mean area is 0.5−0.08=0.42⇒z=1.40.5-0.08=0.42\Rightarrow z=1.4.

  1. Interpret "31% under 45." The area to the left of x=45x=45 is 0.31<0.50.31<0.5, so 45 lies below the mean.

P(0<Z<z)=0.5−0.31=0.19 ⇒ z=0.5P(0<Z<z)=0.5-0.31=0.19\ \Rightarrow\ z=0.5

Since 45 is below the mean, the signed z-value is −0.5-0.5:

45−μσ=−0.5 ⇒ μ−0.5σ=45...(1)\frac{45-\mu}{\sigma}=-0.5\ \Rightarrow\ \mu-0.5\sigma=45 \quad\text{...(1)}

  1. Interpret "8% over 64." The area to the right of x=64x=64 is 0.080.08, so 64 lies above the mean.

P(0<Z<z)=0.5−0.08=0.42 ⇒ z=1.4P(0<Z<z)=0.5-0.08=0.42\ \Rightarrow\ z=1.4

Since 64 is above the mean, the signed z-value is +1.4+1.4:

64−μσ=1.4 ⇒ μ+1.4σ=64...(2)\frac{64-\mu}{\sigma}=1.4\ \Rightarrow\ \mu+1.4\sigma=64 \quad\text{...(2)} …

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