Question

. The saving bank account of a custorner showed an average balance of tRs.150 and a 3D
Rs.50 assuming that the account balances are normally distributed.
i) What % of account is over Rs. 200?
ii) What % of account is is between Rs. 120 and Rs.170?
iii) What % of account is less then Rs.75?​

Answer 1

(i) the percentage of accounts over Rs. 200 = 15.87 %

(ii)The percentage of accounts between Rs. 120 and Rs. 170 = 38.11 %

(iii) The percentage of accounts  less than Rs. 75 = 6.68 %

Given:

mean = μ = 150

standard deviation = σ = 50

To Prove:

i) What % of accounts is over Rs. 200?

ii) What % of the account is between Rs. 120 and Rs.170?

iii) What % of accounts is less than Rs.75?​

Solution:

Given that,

mean = μ = 150

standard deviation = σ = 50

using the normalization distribution,

a ) P (x  > 200 )

= 1 - P (x < 200)

= 1 - P ( x -  μ / σ ) < ( 200 - 150 / 50)

= 1 - P ( z < 50 / 50 )

= 1 - P ( z < 1 )

Using z table

= 1 - 0.8413

= 0.1587

Probability = 0.1587

percentage of accounts over Rs. 200 = 0.1587 * 100

⇒ 15.87%

b ) P (120 < x < 170 )

P ( 120 - 150 / 50) < ( x -  μ / σ ) < ( 170 - 150 / 50)

P ( - 30 / 50 < z < 20 / 50 )

P (-0.60 < z < 0.40 )

P ( z < 0.40 ) - P ( z < -0.60 )

Using z table

= 0.6554 -0.2743

= 0.3811

Probability = 0.3811

Percentage of accounts  between Rs. 120 and Rs. 170 = 0.3811 * 100

⇒ 38.11%

c ) P( x < 75 )

P ( x - μ / σ ) < ( 75 - 150 / 50 )

P ( z < -75 / 50 )

P ( z <-1.5)

= 0.0668

Probability =0.0668

The percentage of accounts  less than Rs. 75 = 0.0668 * 100

⇒6.68%

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