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 1295 1610 1715 1190 1400 1505 Distribution of Prices What is the approximate percentage of buyers who paid less than $1190? 2.5 % What is the approximate percentage of buyers who paid between $1085 and $1400? 4% What is the approximate percentage of buyers who paid between $1190 and $1400? % 47.5 What is the approximate percentage of buyers who paid more than $1715? % 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

B) P(1085 < X < 1400)

= P((1085 - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (X - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (1400 - $\dpi{120} \mu$)/$\dpi{120} \sigma$)

= P((1085 - 1400)/105 < Z < (1400 - 1400)/105)

= P(-3 < Z < 0)

= P(Z < 0) - P(Z < -3)

= 0.5 - 0.0013

= 0.4987

= 49.87%

d) P(X > 1715)

= P((X - $\dpi{120} \mu$)/$\dpi{120} \sigma$ > (1715 - $\dpi{120} \mu$)/$\dpi{120} \sigma$)

= P(Z > (1715 - 1400)/105)

= P(Z > 3)

= 1 - P(Z < 3)

= 1 - 0.9987

= 0.0013

= 0.13%

E) P(1295 < X < 1505)

= P((1295 - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (X - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (1505 - $\dpi{120} \mu$)/$\dpi{120} \sigma$)

= P((1295 - 1400)/105 < Z < (1505 - 1400)/105)

= P(-1 < Z < 1)

= P(Z < 1) - P(Z < -1)

= 0.8413 - 0.1587

= 0.6826

= 68.26%

F) (1295 < X < 1400)

= P((1295 - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (X - $\dpi{120} \mu$)/$\dpi{120} \sigma$ < (1400 - $\dpi{120} \mu$)/$\dpi{120} \sigma$)

= P((1295 - 1400)/105 < Z < (1400 - 1400)/105)

= P(-1 < Z < 0)

= P(Z < 0) - P(Z < -1)

= 0.5 - 0.1587

= 0.3413

= 34.13%