Does a sh script have lower priority than the Real-Time processes which is created by the script in linux?

I have a sh script:

#!/bin/sh
taskset -c 1 ./RR_FIFO_sched 2 90 &
taskset -c 1 ./RR_FIFO_sched 2 90 &
sleep 5s
taskset -c 1 ./RR_FIFO_sched 2 95 &
exit

And the code in the source file named RR_FIFO_sched.c:

#include <sched.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>

int main(int argc, char *argv[])
{
    long i, j, temp;
    char cmd_str[100];
    int rc, current_scheduler_policy;
    struct sched_param my_params;

    if (argc != 3) {
        printf("Usage:RR-FIFO-sched sched_class priority \nsched_class: 0 for CFS; 1 for FIFO; 2 for RR\n");
        exit(1);
    }

    my_params.sched_priority = atoi(argv[2]);
    rc = sched_setscheduler(0, atoi(argv[1]), &my_params);

    if (rc < 0) {
        perror("sched_setscheduler error\n");
        exit(0);
    }

    current_scheduler_policy = sched_getscheduler(0);
    printf("the PID:%d current scheduler = %d \n", getpid(), current_scheduler_policy);
    for (i = 0; i < 1024 * 1024 * 1024; i++)
        for (j = 0; j < 10; j++)
            temp++;
    sprintf(cmd_str, "cat /proc/%d/sched > ./sched-%d ; date >> ./sched-%d",
        getpid(), getpid(), getpid());
    system(cmd_str);
    return 0;
}

I was going to create two RR processes with a priority of 90 at the same time and let them do simple calculations and compete for the CPU#1. Five seconds later, I would create a FIFO process with a priority of 95.In my imagination, the FIFO process will interrupts the first two processes and monopolizes CPU#1. After the FIFO process finishes, the two RR processes regain the CPU. Then with running the script on the terminal and observing with the top command on another terminal, I found that the system created processes one by one instead of creating two processes at the same time. So Why did that happend? I would appreciate it deeply if you would anwser this question for me.