Math, asked by vikasbains98, 7 months ago

An aptitude test for selecting officers in a bank was conducted on 1,000 candidates, the

average score being 42 and the standard deviation being 24. Assuming normal distribution

for the score, find :

a) The number of candidates whose scores exceed 58

b) The number of candidates whose scores lie between 30 and 66.​

Answers

Answered by abiabitha545
0

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Answered by rinayjainsl
0

Answer:

a)The number of candidates whose scores exceed 58 are 252

b)The number of candidates whose scores lie between 30 and 66 are 533

Step-by-step explanation:

Given that,

Number of candidates are n=1000

Average score is N=42

The standard deviation is σ=24

The test statistic is

Z=\frac{X-N}{σ}

(a)The probability of number of candidates whose scores are greater than 50 are

P(X>58)=P(Z>\frac{58-42}{24}) \\ =P(Z>\frac{16}{24})\\=P(Z>0.667) \\ = 1 - P(Z < 0.667)

From normal distribution table we get the above value as

1 - (0.74761) = 0.25239

Therefore number of students with score greater than 50 are

1000 \times 0.25239 = 252.39≈252 \: students

(b)Probability of students whose scores are between 30 and 66 are

P(30<X<66) \\ =P(\frac{30-42}{24}<Z<\frac{66-42}{24})=P(-0.5<Z<1) \\ =P(Z<1)-P(Z<-0.5)

From normal distribution table the above value is calculated as

0.8413 - 0.30854 = 0.53276

Therefore,the number of students with scores in between 30 and 66 are

0.53276 \times 1000 = 532.76 \\ ≈533 \: students

#SPJ3

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