# 1)If a variable X is normally

1)If a variable X is normallydistributed with a mean of 60 and a standard deviation of 15, whatis

P(X > 60)

P(X > 75)

P(X > 80)

P(X < 50)

P(45 < X < 75)

P(X < 45)

Solution :

Given that ,

mean = = 60

standard deviation = = 15

1) P(x > 60) = 1 – p( x< 60)

=1- p P[(x – ) / < (60 – 60) / 15]

=1- P(z < 0.00)

Using z table,

= 1 – 0.5

= 0.5

2) P(x > 75) = 1 – p( x< 75)

=1- p P[(x – ) / < (75 – 60) / 15]

=1- P(z < 1.00)

Using z table,

= 1 – 0.8413

= 0.1587

3) P(x > 80) = 1 – p( x< 80)

=1- p P[(x – ) / < (80 – 60) / 15]

=1- P(z < 1.33)

Using z table,

= 1 – 0.9082

= 0.0918

4) P(x < 50) = P[(x – ) / < (50 – 60) / 15]

= P(z < -0.67)

Using z table,

= 0.2514

5) P( 45 < x < 75) = P[(45 – 60)/ 15) < (x – ) /  <(75 – 60) / 15) ]

= P( -1.00 < z < 1.00)

= P(z < 1.00) – P(z < -1.00)

Using z table,

= 0.8413 – 0.1587

= 0.6826

6) P(x < 45) = P[(x – ) / < (45 – 60) / 15]

= P(z < -1.00)

Using z table,

= 0.1587

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