Question

X is a normally distributed random variable with mean 57 and standard deviation 19.

Answer 1

Answer:

The probability that X is between 38 and 76 is 0.68268.

Step-by-step explanation:

We are given that X is a normally distributed random variable with mean 57 and standard deviation 19.

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

                             Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean = 57

            $\sigma$ = standard deviation = 19

Now, the probability that X is between 38 and 76 is given by = P(38 < X < 76)

         P(38 < X < 76)  =  P(X < 76) - P(X $\leq$ 38)

         P(X < 76) = P( $\frac{X-\mu}{\sigma}$ < $\frac{76-57}{19}$ ) = P(Z < 1) = 0.84134

         P(X $\leq$ 38) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{38-57}{19}$ ) = 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(38 < X < 76) = 0.84134 - 0.15866 = 0.68268.