Answer :
Answer:
33.36% probability that X is less than 2.
Step-by-step explanation:
The distribution is not normal, however, using the central limit theorem, it is going to be approximately normal. So
Central Limit Theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
Normal Probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 2.2, \sigma = 1.4, n = 9, s = \frac{1.4}{\sqrt{9}} = 0.467[/tex]
(a) Suppose we let X be the mean number of accidents per week at the intersection during 9 randomly chosen weeks. What is the probability that X is less than 2?
This is the pvalue of Z when X = 2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2 - 2.2}{0.467}[/tex]
[tex]Z = -0.43[/tex]
[tex]Z = -0.43[/tex] has a pvalue of 0.3336.
33.36% probability that X is less than 2.