Question

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1200 and the standard deviation is $135.

What is the approximate percentage of buyers who paid between $1200 and $1605?
%

What is the approximate percentage of buyers who paid more than $1470?
%

What is the approximate percentage of buyers who paid less than $795?
%

What is the approximate percentage of buyers who paid between $1200 and $1335?
%

What is the approximate percentage of buyers who paid between $1065 and $1335?
%

What is the approximate percentage of buyers who paid between $1200 and $1470?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 1200

standard deviation = $\sigma$ = 135

a) P(1200 < x < 1605) = P[(1200 - 1200)/ 135) < (x - $\mu$) /$\sigma$  < (1605 - 1200) /135 ) ]

= P(0 < z < 3.0)

= P(z < 3.0) - P(z < 0)

Using z table,

= 0.9987 - 0.5

= 0.4987

percent = 49.87%

b) P(x > 1470) = 1 - p( x< 1470)

=1- p P[(x - $\mu$) / $\sigma$ < (1470 - 1200) / 135]

=1- P(z < 2.0)

= 1 - 0.9772

= 0.0228

percent = 2.28%

c) P(x < 795)

= P[(x - $\mu$ ) / $\sigma$ < (795 - 1200) / 135 ]

= P(z < -3.0)

Using z table,

= 0.0013

percent = 0.13%

d) P(1200 < x < 1335) = P[(1200 - 1200)/ 135) < (x - $\mu$) /$\sigma$  < (1335 - 1200) /135 ) ]

= P(0 < z < 1.0)

= P(z < 1.0) - P(z < 0)

Using z table,

= 0.8413 - 0.5

= 0.3413

percent = 34.13%

e) P(1065 < x < 1335) = P[(1065 - 1200)/ 135) < (x - $\mu$) /$\sigma$  < (1335 - 1200) /135 ) ]

= P(-1.0 < z < 1.0)

= P(z < 1.0) - P(z < -1.0)

Using z table,

= 0.8413 - 0.1587

= 0.6826

percent = 68.26%

f) P(1200 < x < 1470) = P[(1200 - 1200)/ 135) < (x - $\mu$) /$\sigma$  < (1470 - 1200) /135 ) ]

= P(0 < z < 2.0)

= P(z < 2.0) - P(z < 0)

Using z table,

= 0.9772 - 0.5

= 0.4772

percent = 47.72%