Question

Find P(-2.85 < z < 1.32) Use the Normal table and give answer using 4 decimal places.

q2. A manufacturer knows that their items have a normally distributed length, with a mean of 17.1 inches, and standard deviation of 2.9 inches.

If one item is chosen at random, what is the probability that it is less than 22.5 inches long?

q3. A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 2.7 years.

If you randomly purchase one item, what is the probability it will last longer than 3 years?

Use the normal table and round answer to four decimal places

q4. In the country of United States of Heightlandia, the height measurements of ten-year-old children are approximately normally distributed with a mean of 54.2 inches, and standard deviation of 3.7 inches.

What is the probability that the height of a randomly chosen child is between 53.15 and 60.95 inches?

Use the Normal table and give answer to 4 decimal places

Answer 1

solution

using z table

(A)P(-2.85 < z < 1.32) =P(Z<1.32) - P(Z < -2.85)

=0.9066- 0.0022

=0.9044

Solution :

Given that ,

mean = = 17.1

standard deviation = = 2.9

n = 1

=17.1

=  / n = 1 / 2.9 = 2.9

P( < 22.5) = P[( - ) / < (22.5 - 17.1) /2.9 ]

= P(z < 1.86)

Using z table

= 0.9686   

(B)Solution :

Given that ,

mean = = 10.1

standard deviation = = 2.7

n = 1

=10.1  

= / n =2.7 / 1 = 2.7

P( >3 ) = 1 - P( < 3)

= 1 - P[( - ) / < (3 - 10.1) /2.7 ]

= 1 - P(z <-2.63 )

Using z table,    

= 1 - 0.0043

= 0.9957

(D)Solution :

Given that,

mean = = 54.2

standard deviation = =3.7

P(53.15< x < 60.95) = P[(53.15 - 54.2) /3.7 < (x -) / < (60.95 - 54.2) / 3.7)]

= P( -0.28< Z <1.82 )

= P(Z <1.82 ) - P(Z <-0.28 )

Using z table,  

= 0.9656 - 0.3897

=0.5759