1. Interpret a 95% confidence interval of (3.355, 3.445) for the population mean. 2. A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni. The mean annual income was $52,500 with a standard deviation of $1,800. Find a 95% confidence interval for the true mean annual income of the nursing school alumni. Write a statement about the confidence level and the interval you find. Solution Given: 1. 95% confidence interval of (3.355, 3.445) for the population mean probability that population mean lies in the interval (3.355,3.445) is 0.95 or you are 95% confident that unknown population parameter (here it is mean) lies in the range given 2.A nursing school wants to estimate the true mean annual income of its alumni. It randomly samples 200 of its alumni mean annual income = $52,500 standard deviation = $1,800. Confidence interval for mean, given sample from population with unknown mean & variance we can calulate t value t= n (xbar - u)/s sample size =200 , therefore d.f of t value =199 CI = ( xbar - t(0.975,199)* s/n , xbar + t(0.975,199)* s/n ) = ( 52500 - 1.975 * 1800/200 , 52500 - 1.975 * 1800/200 ) = ( 52248.62 , 52751.38 ).

1. 1. IQ scores are normally distributed and have a mean of 100 and standard deviation of 15.
a.What is the probability that a randomly selected individual has an IQ of less than 95? b.What
is the probability that a randomly selected individual has an IQ over 130? c.What is the
probability that a randomly selected individual has an IQ between 90 and 120? d.What
proportion of the population has an IQ between 85 and 115? Between 70 and 130? e.Would it
be unusual for an individual to have an IQ over 140? f.The membership requirement for Mensa
is that one must have an IQ that is in the top 2% (www.mensa.org). What is the minimum score
required for Mensa membership? g.An individual claims that he know three people with IQs
over 200. Would you believe his claim? Why or why not?
Solution
(a) P(X<95) = P((X-mean)/s <(95-100)/15)
=P(Z<-0.33) = 0.3707 (from standard normal table)
(b) P(X>130) = P(Z>(130-100)/15) = P(Z>2) =0.0228
(c) P(90140) = P(Z>(140-100)/15)
=P(Z>2.67) = 0.0038
Since the probability is small, it would be unusual for an individual to have an IQ over 140
(f) P(X>c)=0.02
--> P(Z<(c-100)/15)=1-0.02=0.98
--> (c-100)/15 =2.05 (from standard normal table)
--> c= 100+2.05*15 =130.75
(g) The test hypothesis is
Ho:mu=200
Ha: mu>200
The test statistic is
Z=(xbar-mu)/(s/vn)
=(200-100)/(15/sqrt(3))
=11.55
The p-value is P(Z>11.55)=0 (from standard normal table)
Since the p-value is 0, we reject Ho.
We can conclude that we believe his claim.