Question

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1400 and the standard deviation is $105. + 1085 1190 1295 1400 1505 1610 1715 Distribution of Prices What is the approximate percentage of buyers who paid less than $1190? % What is the approximate percentage of buyers who paid between $1085 and $1400? at% What is the approximate percentage of buyers who paid between $1190 and $1400? % What is the approximate percentage of buyers who paid more than $1715? x*% What is the approximate percentage of buyers who paid between $1295 and $1505? % What is the approximate percentage of buyers who paid between $1295 and $1400? x% Get help: Video Video

Answer 1

1. The approximate percentage of buyer who paid less than 1190 is calculated as follows, P(X<1190) = P( 854 119013.400 = P(Z <-2) 2.5% The required percentage is obtained from standard normal distribution table. 2. The approximate percentage of buyer who paid between 1085 and 1400 is calculated as follows, 1400-1400 105 ) P(1085<X <1400)= P 1085-1400 X- 1050 =P(-3<Z<0) = 0.5-0.00135 0.5 or 50% The required percentage is obtained from standard normal distribution table.

3. The approximate percentage of buyer who paid between 1190 and 1400 is calculated as follows, X-u 1400-1400 105 ) 1190-1400 P(1190 < X <1400) 105C = P(-2<Z<0) = 0.5-0.025 0.475 or 47.5% The required percentage is obtained from standard normal distribution table. 4. The approximate percentage of buyer who paid more than 1715 is calculated as follows, P(X>1715)=P(X=4, 1715–2400) = P(Z >3) 0.00135 or 0.135% The required percentage is obtained from standard normal distribution table.