Question #220491

a sample of 600 respondents the monthly income of the respondents follow a normal distribution with its mean and standard deviation at 30,000 and 6000 respectively
What is the probability that monthly income is between 20000 and 34000
Probability that a monthly income is more than 32,000
How many people Expected to have the income between 20,000 and 34000
What is the probability that the monthly income is less than 24,000

Expert's answer

N=600μ=30000σ=6000N=600 \\ \mu=30000 \\ \sigma = 6000

Is it justifiable to exclude homemakers’ value creation from national income?

P(20000<X<34000)=P(X<34000)−P(X<20000)=P(Z<34000−300006000−P(Z<20000−300006000)=P(Z<0.666)−P(Z<−1.666)=0.7476−0.0478=0.6998P(20000<X<34000) = P(X<34000) -P(X<20000) \\ = P(Z< \frac{34000-30000}{6000} -P(Z< \frac{20000-30000}{6000}) \\ =P(Z< 0.666) -P(Z< -1.666) \\ = 0.7476 -0.0478 \\ = 0.6998

Probability that a monthly income is more than 32,000

P(X>32000)=1−P(X<32000)=1−P(Z<32000−300006000)=1−P(Z<0.3333)=1−06304=0.3696P(X>32000) = 1 -P(X<32000) \\ = 1 -P(Z< \frac{32000-30000}{6000}) \\ = 1 -P(Z< 0.3333) \\ = 1 -06304 \\ = 0.3696

How many people Expected to have the income between 20,000 and 34000

Number of people =N×P(20000<X<34000)= N \times P(20000<X<34000)

=600×0.6998=419.88≈420= 600 \times 0.6998 \\ = 419.88 ≈420

What is the probability that the monthly income is less than 24,000

P(X<24000)=P(Z<24000−300006000)=P(Z<−1)=0.1586P(X< 24000) = P(Z< \frac{24000-30000}{6000}) \\ = P(Z< -1) \\ = 0.1586


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