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Question:
Suppose the amount of sun block lotion in plastic bottles leaving a filling machine has a normal distribution. The bottles are labeled 300 milliliters (ml) but the actual mean is 302 ml and the standard deviation is 2 ml. If you purchase a package of 6 bottles of lotion, what is closest to the probability that at least one bottle has a content of less than 300 ml?
Answer:
0.645314
Step-by-step explanation:
Given:
Mean, u = 302
Standard deviation = 2
n = 6
Let's first find P(X>300):
Using the standard normal table,
NORMSDIST(-1) = 0.158655
Thus,
P(Z<-1) = 0.158655
Find the closest to the probability that at least one bottle has a content of less than 300 ml:
Given that 6 bottles were purchased & p = 0.158655
Find:
P(X≥1) = 1 – P(X<1) = 1 – P(X=0)
Use bimonial distribution:
Therefore,
P(X≥1) = 1 – P(X<1) = 1 – P(X=0)
= 1 - 0.354686
= 0.645314
The closest to the probability that at least one bottle has a content of less than 300 ml is 0.645314
Using the t-distribution, the 96% confidence interval is given as follows:
(18.96, 21.04).
<h3>What is a t-distribution confidence interval?</h3>The confidence interval is:
In which:
is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 96% confidence interval, with 100 - 1 = 99 df, is t = 2.0812.
The parameters are given as follows:
Hence the bounds of the interval are:
More can be learned about the t-distribution at brainly.com/question/16162795
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