Try to use y=Mx+b and if that doesn’t help just say so and I will help even more
Answer:
c). Two tailed test
Step-by-step explanation:
The given hypothesis are
Null hypothesis: H0:μ= 1.7
Alternative hypothesis: H1:μ≠ 1.7
The alternative hypothesis demonstrates that mean number of children are not 1.7 in 2000. This means that mean number of children can be greater than 1.7 or mean number of children can be less than 1.7. Thus, the given alternative hypothesis indicates the two tailed test.
Answer:
a) 0.778
b) 0.9222
c) 0.6826
d) 0.3174
e) 2 drivers
Step-by-step explanation:
Given:
Sample size, n = 5
P = 40% = 0.4
a) Probability that none of the drivers shows evidence of intoxication.



b) Probability that at least one of the drivers shows evidence of intoxication would be:
P(X ≥ 1) = 1 - P(X < 1)
c) The probability that at most two of the drivers show evidence of intoxication.
P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)
d) Probability that more than two of the drivers show evidence of intoxication.
P(x>2) = 1 - P(X ≤ 2)
e) Expected number of intoxicated drivers.
To find this, use:
Sample size multiplied by sample proportion
n * p
= 5 * 0.40
= 2
Expected number of intoxicated drivers would be 2