Question

Recently, the top web browser had 51.66 of the market. In a random sample of 225 people, what is the probability that fewer than 99 did not use the top web browser? Round the final answer to at least 4 decimal places and intermediate -value calculations to decimal places.

Answer 1

Solution :

Given

Sample size , n =    225
Probability of an event of interest, p = 1-0.5166 = 0.4834   
  
Mean = np = 225*0.4834 =108. 765   
std dev ,σ=√np(1-p)=√225*0.4834*(1-0.4834) = 7.4959                      
                          
P(X <   99 ) = P(Xnormal < 98.5   )          
                          
Z=(Xnormal - µ ) / σ = ( 98.5   - 108.765 ) /   7.4959   =   -1.369
                          
=P(Z<   -1.369 ) =    0.0854996

The required probability is

p(x<99) =0.0855

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