Question

The numbers illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1000 and the standard deviation is $60.

820, 880, 940, 1000, 1060, 1120, 1180, are the Distribution of Prices

What is the approximate percentage of buyers who paid between $820 and $1000?
? %

What is the approximate percentage of buyers who paid between $940 and $1000?
? %

What is the approximate percentage of buyers who paid more than $1120?
? %

What is the approximate percentage of buyers who paid between $1000 and $1120?
? %

What is the approximate percentage of buyers who paid between $940 and $1060?
? %

What is the approximate percentage of buyers who paid less than $820?
? %

Answer 1

a) P(820 < X < 1000)

= P(-3 < Z < 0)

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

= 0.5 - 0.0013

= 0.4987

b) P(940 < X < 1000)

= P(-1 < Z < 0)

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

= 0.5 - 0.1587

= 0.3413

c) P(X > 1120)

= P(Z > 2)

= 1 - P(Z < 2)

= 1 - 0.9772

= 0.0228

d) P(1000 < X < 1120)

= P(0 < Z < 2)

= P(Z < 2) - P(Z < 0)

= 0.9772 - 0.5

= 0.4772

e) P(940< X < 1060)

= P(-1 < Z < 1)

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

= 0.8413 - 0.1587

= 0.6826

f) P(X < 820)

= P(Z < -3)

= 0.0013