5.4.2 Probability Distributions, Long Questions

Question 5:

The diameter of oranges harvested from a fruit orchard has a normal distribution with a mean of 3.2 cm and a variance of 2.25 cm.

Calculate

(a) the probability that an orange chosen at random from this fruit orchard has a diameter of more than 3.8 cm.

(b) the value of k if 30.5 % of the oranges have diameter less than k cm.

Solution:

µ = 3.2 cm

σ²= 2.25cm

σ = √2.25 = 1.5 cm

Let X represents the diameter of an orange.

X ~ N (3.2, 1.5²)

(a)

\begin{array}{l} P (X > 3.8) \\ = P (Z > \frac{3.8 - 3.2}{1.5}) \\ = P (Z > 0.4) \\ = 0.3446 \end{array}

(b)

\begin{array}{l} P (X < k) = 0.305 \\ P (Z < \frac{k - 3.2}{1.5}) = 0.305 \\ From the standard normal distribution table, \\ P (Z > 0.51) = 0.305 \\ P (Z < - 0.51) = 0.305 \\ \frac{k - 3.2}{1.5} = - 0.51 \\ k - 3.2 = - 0.765 \\ k = 2.435 \end{array}

Question 6:

The masses of tomatoes in a farm are normally distributed with a mean of 130 g and standard deviation of 16 g. Tomato with weight more than 150 g is classified as grade ‘A’.

(a) A tomato is chosen at random from the farm. Find the probability that the tomato has a weight between 114 g and 150 g.

(b) It is found that 132 tomatoes in this farm are grade ‘A’. Find the total number of tomatoes in the farm.

Solution:

µ = 130

σ = 16

(a)

\begin{array}{l} P (114 < X < 150) \\ = P (\frac{114 - 130}{16} < Z < \frac{150 - 130}{16}) \\ = P (- 1 < Z < 1.25) \\ = 1 - P (Z > 1) - P (Z > 1.25) \\ = 1 - 0.1587 - 0.1056 \\ = 0.7357 \end{array}

(b)

Probability of getting grade ‘A’ tomatoes,

P (X > 150) = P (Z > 1.25)

= 0.1056

\begin{array}{l} Lets total number of tomatoes = N \\ 0.1056 \times N = 132 \\ N = \frac{132}{0.1056} \\ N = 1250 \end{array}

Question 7:
In a boarding school entry exam, 300 students sat for a mathematics test. The marks obtained follow a normal distribution with a mean of 56 and a standard deviation of 8.

(a) Find the number of students who pass the test if the passing mark is 40.

(b) If 12% of the students pass the test with grade A, find the minimum mark to obtain grade A.

Solution:

\begin{array}{l} Let X = marks obtained by students \\ X ~ N (56, 8^{2}) \\ (a) P (X \geq 40) = P (Z \geq \frac{40 - 56}{8}) \\ = P (Z \geq - 2) \\ = 1 - P (Z \geq 2) \\ = 1 - 0.02275 \\ = 0.9773 \\ Number of students who pass the test \\ = 0.9773 \times 300 \\ = 293 \\ (b) Let the minimum mark to obtain grade A be k \\ P (X \geq k) = 0.12 \\ P (Z \geq \frac{k - 56}{8}) = 0.12 \\ \frac{k - 56}{8} = 1.17 \\ k = (1.17) (8) + 56 \\ = 65.36 \end{array}

Thus, the minimum mark to obtain grade A is 66.

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