Answer:
C. 0.1515
Step-by-step explanation:
The main objective here is to find the P-value for the test of [tex]H_0[/tex]
Given that ;
the mean value = 2.2
the standard deviation = 1.4
number of recorded accident per week = 52
The null hypothesis is : [tex]H_o: \mu =2[/tex]
The alternative hypothesis is : [tex]H_A = \mu < 2[/tex]
The Z- value can be calculated as:
[tex]z = \dfrac{x- \mu}{\dfrac{\sigma }{\sqrt{n}}}[/tex]
[tex]z = \dfrac{2- 2.2}{\dfrac{1.4 }{\sqrt{52}}}[/tex]
[tex]z = \dfrac{- 0.2}{\dfrac{1.4 }{7.211}}[/tex]
z = -1.03
From the normal distribution table for probability;
P(z< -1.03 ) = 0.1515