Answer: 0.819
Step-by-step explanation:
Given: X is a normally distributed random variable with mean 95 and standard deviation 3.
Then, the probability that X is between 89 and 98 will be:
[tex]P(89<X<95)=P(\dfrac{89-95}{3}<\dfrac{X-mean}{standard\ deviation}<\dfrac{98-95}{3})\\\\=P(-2<Z<1)\ \ \ [Z=\dfrac{X-mean}{standard\ deviation}]\\\\ =P(Z<1)-(1-P(Z<2))\\\\= 0.8413- (1-0.9772)\ \ \ [\text{By p-value table}]\\\\= 0.8185\approx0.819[/tex]
Hence, the required probability = 0.819
