Question

A company’s cereal boxes advertise that each box contains 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean μ = 9.70 ounces and standard deviation σ = 0.03 ounce. Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal in these boxes is less than 9.65 ounces?
a. 0.4525
b. 0.9522
c. 0.9999
d. 0.0478
e. 0.0001

Answer 1

Answer:

Answer: 0.0001

Step-by-step explanation:

Given

Mean μ=9.70

Sample size=n=5

standard deviation=σ = 0.03

Since its a normal distribution,so

σx=σ/$\sqrt{n}$

σx=0.03/$\sqrt{5}$

σx=0.0134

Now in order to use the table we have to to figure out z value.

z=(x-μ)/σ

z=(9.65-9.70)/0.0134

z=-3.73

From the probability index tables

So P(z≤-3.73)=0.0001

Answer 2

The probability that the mean amount of cereal in these boxes is less than 9.65 ounces is 0.0001

Given that,  

•  Mean μ=9.70
• Sample size=n=5
• standard deviation=σ = 0.03

Since its a normal distribution,so

$\sigma x=\sigma/\\\\\sigma x=0.03\\\\\sigma x=0.0134$  

Now in order to use the table we have to to figure out z value.

$z=(x-\mu)\div\sigma\\\\z=(9.65-9.70)\div 0.0134$

z=-3.73

Now

From the probability index tables

So P(z≤-3.73)=0.0001

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