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Q.Given that the scores of a set of candidates on an IQ test are normally distributed. If the IQ test has a mean of 100 and a standard deviation of 10, determine the probability that a candidate who takes the test will score between 90 and 110. [Given P(Z<1)=0.8413P(Z < 1) = 0.8413 and P(Z<−1)=0.1587P(Z < -1) = 0.1587]

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★est
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Standardising, 9090 and 110110 become z=−1z=-1 and z=+1z=+1; P(−1<Z<1)=0.8413−0.1587=0.6826P(-1<Z<1)=0.8413-0.1587=0.6826.

Standard normal variable Z=X−μσZ=\dfrac{X-\mu}{\sigma}, where μ\mu is the mean and σ\sigma the standard deviation. Then P(a<X<b)=P(Z<zb)−P(Z<za)P(a<X<b)=P(Z<z_b)-P(Z<z_a).

Given μ=100, σ=10\mu=100,\ \sigma=10; find P(90<X<110)P(90<X<110).

  1. Standardise the lower limit: z1=90−10010=−1z_1=\dfrac{90-100}{10}=-1.
  2. Standardise the upper limit: z2=110−10010=+1z_2=\dfrac{110-100}{10}=+1. …

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