Answer:
C. 0.1515
Explanation:
The main objective here is to find the P-value for the test of
![H_0](https://img.qammunity.org/2021/formulas/mathematics/college/ukwsil2snwe4lwpu4rcvgvyifselbouisn.png)
Given that ;
the mean value = 2.2
the standard deviation = 1.4
number of recorded accident per week = 52
The null hypothesis is :
![H_o: \mu =2](https://img.qammunity.org/2021/formulas/mathematics/college/29v6skibjoswfrqfuho3s6flk7mhgspe5w.png)
The alternative hypothesis is :
![H_A = \mu < 2](https://img.qammunity.org/2021/formulas/mathematics/college/gc3ttexdm9it850t5l9ch2coocnpj1p6p6.png)
The Z- value can be calculated as:
![z = (x- \mu)/((\sigma )/(√(n)))](https://img.qammunity.org/2021/formulas/mathematics/college/l0lu5gmfqsli25vc9f791rii9ncau10kqs.png)
![z = (2- 2.2)/((1.4 )/(√(52)))](https://img.qammunity.org/2021/formulas/mathematics/college/ivby8f7p0djq1jcyjxpgqv61nekl6jo3db.png)
![z = (- 0.2)/((1.4 )/(7.211))](https://img.qammunity.org/2021/formulas/mathematics/college/9fpm5w9p321fuaon0furjkhnz6qzbm5q94.png)
z = -1.03
From the normal distribution table for probability;
P(z< -1.03 ) = 0.1515