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Check Your Progress · Q23

Q.In an examination, 2000 students appeared and the mean of the normal distribution of marks is 30 with standard deviation as 6.25. Find out how many students are expected to score.
i. between 20 and 40 marks.
ii. less than 25 marks

Puducherry CbseNCERTSubjective· 3mImportance★★★★★est
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Standardise each mark with z=x−μσz=\dfrac{x-\mu}{\sigma} (μ=30, σ=6.25\mu=30,\ \sigma=6.25), read the normal-table areas, then multiply the probabilities by the 2000 students. Result: about 1781 score between 20 and 40, and about 424 score under 25.

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

Expected count: (number)=N×P(event)\text{(number)}=N\times P(\text{event})

where xx = a mark, μ=30\mu=30 = mean, σ=6.25\sigma=6.25 = standard deviation, N=2000N=2000 = total students. Table values used: P(0<Z<1.6)=0.4452P(0<Z<1.6)=0.4452, P(0<Z<0.8)=0.2881P(0<Z<0.8)=0.2881.

(i) Between 20 and 40 marks

  1. Convert the lower limit x=20x=20:

z1=20−306.25=−106.25=−1.6z_1=\frac{20-30}{6.25}=\frac{-10}{6.25}=-1.6

  1. Convert the upper limit x=40x=40:

z2=40−306.25=106.25=+1.6z_2=\frac{40-30}{6.25}=\frac{10}{6.25}=+1.6

  1. Required probability (area between z=−1.6z=-1.6 and z=1.6z=1.6, symmetric about the mean):

P(−1.6<Z<1.6)=2×P(0<Z<1.6)=2×0.4452=0.8904P(-1.6<Z<1.6)=2\times P(0<Z<1.6)=2\times0.4452=0.8904

  1. Expected number of students:

N×P=2000×0.8904=1780.8≈1781 studentsN\times P=2000\times0.8904=1780.8\approx 1781\text{ students}

(ii) Less than 25 marks

  1. Convert x=25x=25:

z=25−306.25=−56.25=−0.8z=\frac{25-30}{6.25}=\frac{-5}{6.25}=-0.8

  1. Required probability (area to the left of z=−0.8z=-0.8): P(Z<−0.8)=0.5−P(0<Z<0.8)=0.5−0.2881=0.2119P(Z<-0.8)=0.5-P(0<Z<0.8)=0.5-0.2881=0.2119 …
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