Question

The random variable X, which represents the number of cherries on a cake, has the following probability distribution:

x	f(x)
4	0.2
5	0.4
6	0.3
7	0.1

Calculate the probability that the average number of cherries in 36 cakes is less than 5.1

Write the answer with 4 decimals.

Answer 1

Solution:

x P(X) X.p(X) 4 0.2 4.0.2 = 0.8 LO 0.4 5.0.4 = 2 6 0.3 6.0.3= 1.8 7 0.1 7.0.1 = 0.7 Therefore, the population mean is calculated as follows: X p(X;) = 4.0.2 + 5.0.4+6.0.3+7.0.1 = 5.3 Now, we need to multiply the corresponding squares XP of the outcomes with the corresponding probabilities, in order to compute the population variance o: p(X) XpX) X.p(X) 42.0.2 = 3.2 4 0.2 4.0.2 = 0.8 0.4 5.0.4 = 2 52.0.4 = 10 5 6 0.3 6.0.3 = 1.8 62.0.3 = 10.8 7 0.1 7.0.1 = 0.7 74.0.1 = 4.9 Therefore, we first compute the expected value of X2: E(X) - X?p(X) 42.0.2 +52.0.4+62.0.3+72.0.1 28.9

Therefore, the population variance is computed as follows: E(X) - E(X) 28.9-5.32 0.81 And finally, taking square root to the variance we get that the population standard deviation is o = Vol = V0.81 = 0.9. Therefore, based on the data provided for the discrete distribution, the population mean is u = 5.3 and the population standard deviation is o = 0.9

probability that the average number of cherries in 36 cakes is less than 5.1

σ_x = 0.9 / Sqrt(36) = 0.15

Since 5.3 and o = 0.15 we have: P(X<5.1)=P(X-x<5.1 -5.3)=P PA*** 5.1-5.3 0.15 5.1-5.3 Z and Since -1.33 we have 0.15 P(X<5.1)=P(Z <-1.33) Step 3: Use the standard normal table to conclude that: P(Z < -1.33) = 0.0918