The probability that a sample of size n of a normally distributed data with mean, μ and standard deviation, σ, is between two values, a and b, is given by
[tex]P(a\ \textless \ x\ \textless \ b)=P(x\ \textless \ b)-P(x\ \textless \ a) \\ \\ =P\left(z\ \textless \ \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma/\sqrt{n}} \right)[/tex]
Given that the
intelligence quotient (iq) test scores for adults are normally
distributed with a mean of 100 and a standard deviation of 15.
The probability that a sample of 50 adults will have a
mean of between 98 and 103 is given by:
[tex]P(98\ \textless \ x\ \textless \ 103)=P(x\ \textless \ 103)-P(x\ \textless \ 98) \\ \\ =P\left(z\ \textless \ \frac{103-100}{15/\sqrt{50}} \right)-P\left(z\ \textless \ \frac{98-100}{15/\sqrt{50}} \right) \\ \\ =P\left(z\ \textless \ \frac{3}{2.121} \right)-P\left(z\ \textless \ \frac{-2}{2.121} \right) \\ \\ =P(z\ \textless \ 1.414)-P(z\ \textless \ -0.9428) \\ \\ =0.92135-0.17289=0.74846[/tex]
Therefore, the probability that a sample of 50 adults will have a
mean of between 98 and 103 is 0.74846
