Question

A geneticist working for a seed company develops a new carrot for growing in heavy clay soil. After measuring 5,000 of these carrots, it can be said that carrot length, X, is normally distributed with a mean of 11.5 cm and a standard deviation of 1.15 cm. What is the probability that X (a carrot’s length), will take on a value in the interval 10.0 cm ≤ x ≤ 13.0 cm?

What is the probability of harvesting a carrot of less than or equal to 13.5 cm? If two carrots are harvested, presumably independent events, what then is the probability that both of them will be longer than 13.5 cm?

Answer 1

Solution :

Given that ,

mean = = 11.5

standard deviation = = 1.15

a) P( 10< x <13 ) = P[(10 - 11.5)/ 1.15) < (x - ) /  < (13 -11.5) /1.5 ) ]

= P( -1< z < 1 )

= P(z < 1 ) - P(z < -1)

Using standard normal table

= 0.8413 -0.1587 = 0.6826

Probability = 0.6826

b)

P(x < 13.5) = P[(x - ) / < (13.5 -11.5) /1.5 ]

= P(z < 1.33)

= 0.9082

probability =0.9082

c)

n = 2

= = 11.5

= / n = 1.15/ 2 = 0.7778

P( > 13.5) = 1 - P( < 13.5)

= 1 - P[( - ) / < (13.5 -11.5) /0.7778 ]

= 1 - P(z < 2.57 )

= 1 - 0.9949 = 0.0051

Probability = 0.0051