Question

The mean annual cost of car insurance is Rs 9450 and standard deviation of Rs 250. What is the probability that the average insurance cost of a sample of 25 cars will be between Rs 9400 and Rs 9500?

Answer 1

The probability that the average insurance cost of a sample of 25 cars will be between Rs 9400 and Rs 9500 is 0.6826

Given :

The mean annual cost of car insurance is Rs 9450 and standard deviation of Rs 250

To find :

The probability that the average insurance cost of a sample of 25 cars will be between Rs 9400 and Rs 9500

Solution :

Step 1 of 2 :

Write down mean and standard deviation

Here it is given that the mean annual cost of car insurance is Rs 9450 and standard deviation of Rs 250

Mean = μ = 9450

Standard deviation = σ = 250

Sample size = n = 25

Step 2 of 2 :

Calculate the required probability

$\displaystyle \sf{ z = \frac{X - \mu}{ \dfrac{ \sigma}{ \sqrt{n} } } }$

$\displaystyle \sf{ \implies z = \frac{X - 9450}{ \dfrac{ 250}{ \sqrt{25} } } }$

$\displaystyle \sf{ \implies z = \frac{X - 9450}{ \dfrac{ 250}{5 } } }$

$\displaystyle \sf{ \implies z = \frac{X - 9450}{ 50 } }$

$\displaystyle \sf For \: \: x = 9400 \: \: we \: have \: \: z = \frac{9400 - 9450}{ 50 } = \frac{ - 50}{50} = - 1$

$\displaystyle \sf For \: \: x = 9500 \: \: we \: have \: \: z = \frac{9500 - 9450}{ 50 } = \frac{ 50}{50} = 1$

Hence the required probability

$\displaystyle \sf{ = P\bigg( \: \frac{9400 - 9450}{ \dfrac{ 250}{ \sqrt{25} } } < z < \frac{9500 - 9450}{ \dfrac{ 250}{ \sqrt{25} } }\bigg) }$

$\displaystyle \sf{ } = P( - 1 < z < 1)$

$\displaystyle \sf{ } = 2P( 0 < z < 1)$

$\displaystyle \sf{ = 2 \times 0.3413 }$

$\displaystyle \sf{ = 0.6826 }$

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