Answer: 0.9817
Step-by-step explanation:
Given : The Intelligence Quotient (IQ) test scores for adults are normally distributed with
Mean : [tex]\mu=100[/tex]
Standard deviation : [tex]\sigma=15[/tex]
Sample size : = 50
Let x be the random variable that represents the IQ test scores for adults.
Z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x =95
[tex]z=\dfrac{95-100}{\dfrac{15}{\sqrt{50}}}\approx-2.36[/tex]
For x =105
[tex]z=\dfrac{105-100}{\dfrac{15}{\sqrt{50}}}\approx2.36[/tex]
By using standard normal distribution table , the probability the mean of the sample is between 95 and 105 :-
[tex]P(95<X<105)=P(-2.36<z<2.36)=1-2(P(z<-2.36))\\\\=1-2(0.0091375)=0.981725\appprox0.9817[/tex]
