Question

If X is a normally distributed random variable with mean 14 and standard deviation 2 then P (12 < X < 16) is

Answer 1

P(12 < X < 16) = 0.84134 - 0.15866 = 0.68268.

Step-by-step explanation:

We are given that X is a normally distributed random variable with mean 14 and standard deviation 2.

Let X = a random variable

So, X ~ Normal($\mu=14,\sigma^{2} =2^{2}$)

The z score probability distribution for normal distribution is given by;

                                Z  =  $\frac{X-\mu}{\sigma}$

where, $\mu$ = population mean = 14

            $\sigma$ = standard deviation = 2

Now, Probability that the random variable X is between 12 and 16 is given by = P(12 < X < 16)

         P(12 < X < 16) = P(X < 16) - P(X $\leq$ 12)

     P(X < 16) = P( $\frac{X-\mu}{\sigma}$ < $\frac{16-14}{2}$ ) = P(Z < 1) = 0.84134

     P(X $\leq$ 12) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{12-14}{2}$ ) = P(Z $\leq$ -1) = 1 - P(Z < 1)  

                                                  = 1 - 0.84134 = 0.15866

The above probability is calculated by looking at the value of x = 1 in the z table which has an area of 0.84134.

Therefore, P(12 < X < 16) = 0.84134 - 0.15866 = 0.68268.

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