Analysis of survey data - Pls help

[jsmith241182]

I have two questions to answer which im struggling with, could anyone pls help? the questions are:

1) In a market research project a local authority has requested the evaluation of the demand for a library. Visitors to the library stay for a mean of 3 hours with a standard deviation of 1.8 hours. Within a sample of 1000 visitors how many can be expected to visit the library for (a) less than 2 hours (b) more than 5 hours (c) between 2 and 5 hours. Assume that the distribution is normally distributed.

2) In order to estimate the percentage of dwellings which have a smoke alarm in the northwest to within 1% for a 95% confidence level a pilot survey was carried out. From a random sample of 30 dwellings, 12 were found to have a smoke alarm installed. How many houses need to be sampled in the main survey?

Thanks!

[Rassis]

Question 1:

Mean = 3; SD = 1.8; n = 1,000

P(x <= 2) = 0.28926 or 1,000 x 0.28926 = 289 visitors
P(x > 5) = 0.13326 or 1,000 x 0.13326 = 133 visitors
P(2 <= x <= 5) = 0.86674 - 0.28926 = 0.57748 or 1,000 x 0.57748 = 578 visitors

Question 2:

p = 12/40 = 40%; e = 0.01; CL = 95%

n0 = Z(a/2).p(1 – p)/e^2 = 1,95996 x 0.4 x (1 – 0.4)/0.01^2 = 9,220

If you know the size N of the population, then:

n = (n0).(N)/[n0 + (N – 1)]

If, for instance N = 1,000, the result would be n = 902 instead.