Answer:
a) 0.0167
b) 0
c) 5.948
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 6.16 ounces
Standard Deviation, σ = 0.08 ounces
We are given that the distribution of fill volumes of bags is a bell shaped distribution that is a normal distribution.
Formula:
a) Standard deviation of 23 bags

b) P( fill volume of 23 bags is below 5.95 ounces)
P(x < 5.95)
Calculation the value from standard normal z table, we have,
c) P( fill volume of 23 bags is below 6 ounces) = 0.001
P(x < 6) = 0.001
Calculation the value from standard normal z table, we have,


If the mean will be 5.948 then the probability that the average of 23 bags is below 6.1 ounces is 0.001.
Answer:
a₇ = 2.375
Step-by-step explanation:
There is a common ratio r between consecutive terms, that is
r =
=
=
= - 
This indicates the sequence is geometric with nth term
= a₁ 
where a₁ is the first term and r the common ratio
Here a₁ = 152 and r = -
, then
= 152
, so
a₇ = 152
= 152 ×
= 2.375
Answer:
Answer: 9
Step-by-step explanation:

• For f(0), substitute x with zero:

Answer:
Step-by-step explanation:
Straight angle is equal to 180°
Angle x, 60° and 80° form a straight angle.
<u>Then we have following equation to solve for x:</u>
- x + 60° + 80° = 180°
- x + 140° = 180°
- x = 180° - 140°
- x = 40°
Answer:
Step-by-step explanation:
<em>Given that X - the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.</em>
<em />
<em>Height of male pilot = 74.2 inches</em>
<em />
<em>We have to find the percentile</em>
<em />
<em>X = 74.2</em>
<em />
<em>Corresponding Z score = 74.2-72.6 = 1.6</em>
<em />
<em>P(X<174.2) = P(Z<1.6) = 0.5-0.4452=0.0548=5.48%</em>
<em />
<em>i.e. only 5% are below him in height.</em>
<em />
<em>Thus the malepilot is in 5th percentile.</em>