91Ó°ÊÓ

There are 100 randomly selected cars. There are five types of colors in the sample of cars which are white, black, gray, silver, and red.

Requirements for normal approximation of binomial distribution:

The sample is a simple random sample with size n from a population with a proportion of success p or probability of success p.\(np \ge 5,nq \ge 5.\)

Let X be the random variable for the number of gray cars in a sample of 100.

The proportion of cars with gray color is 0.16(p).

The sample size and probability of success are;

\(\begin{aligned}{l}n = 100\\p = 0.16\end{aligned}\)

The requirement for the normal approximation to binomial distribution, that is, \(np \ge 5,nq \ge 5.\)

Thus,

\(\begin{aligned}{c}np = 100 \times 0.16\\ = 16\\ \ge 5\end{aligned}\)

\(\begin{aligned}{c}nq = n \times \left( {1 - p} \right)\\ = 100 \times \left( {1 - 0.16} \right)\\ = 84\\ \ge 5\end{aligned}\)

Thus, the approximation to normal distribution is verified.

To approximate binomial distribution into normal, it is important to find the mean and standard deviation.

Hence, the formula for mean and standard deviation is given below,

\(\begin{aligned}{l}\mu = n \times p\\\sigma = \sqrt {n \times p \times q} \end{aligned}\)

Now, substitute all the values in the formula,

\(\begin{aligned}{c}\mu = n \times p\\ = 100 \times 0.16\\ = 16\end{aligned}\)

\(\begin{aligned}{c}\sigma = \sqrt {n \times p \times q} \\ = \sqrt {100 \times 0.16 \times 0.84} \\ = 3.6661\end{aligned}\)

Now, the mean of the normal distribution is \(\mu = 16\) and the standard deviation of normal distribution is \(\sigma = 3.6661\).

The probability that the number of gray cars is equal to 10 is expressed as,

\(\begin{aligned}{c}P\left( {X = 10} \right) = P\left( {9.5 < X < 10.5} \right)\\ = P\left( {X < 10.5} \right) - P\left( {X < 9.5} \right)\end{aligned}\)

The formula for z-score is,

\(z = \frac{{x - \mu }}{\sigma }\)

The z-score associated with each value are,

\(\begin{aligned}{c}{z_1} = \frac{{{x_1} - \mu }}{\sigma }\\ = \frac{{10.5 - 16}}{{3.6661}}\\ = - 1.500\end{aligned}\)

\(\begin{aligned}{c}{z_2} = \frac{{{x_2} - \mu }}{\sigma }\\ = \frac{{9.5 - 16}}{{3.6661}}\\ = - 1.773\end{aligned}\)

From the standard normal table,

the cumulative probability for the value -1.5 is 0.0668.

the cumulative probability for the value -1.8 is 0.0359.

Thus, probability that the number of red cars is 14,

\(\begin{aligned}{c}P\left( {X = 10} \right) = P\left( { - 1.8 < Z < - 1.5} \right)\\ = P\left( {Z < - 1.5} \right) - P\left( {Z < - 1.8} \right)\\ = 0.0668 - 0.0359\\ = 0.0309\end{aligned}\)

Thus, the probability that the number of gray cars is 10 is 0.0309.

Significantly high events are those which have probability 0.05 or less for occurrence of events greater than or equal to the given event.

Here, the probability is stated for 10 gray cars and not at least 10 such cars.

Thus, if 10 gray cars is significantly high number or not cannot be determined using the result.