Answer to Question #150742 in Statistics and Probability for abed

Question #150742
A candy company has calibrated their machines to fill each bag of candy to an average weight of 505 grams with a standard deviation of 4 grams. The company tests the consistency of their machines by taking a sample of 25 bags and weighing them.



a- What is the probability that the average weight of candy bags in the sample is less than 503 grams?

b- What is the probability that the sample average is between 503 and 506?

c- Below what value do 95% of the sample means fall?
1
Expert's answer
2020-12-15T02:36:50-0500

μ=505\mu=505

σ=4\sigma=4

n=25n=25

Let's denote xˉ\bar x sample mean

a) P(xˉ<503)=P(Z<503μσn)=P(Z<503505425)=P(Z<2.5)=0.0062P(\bar x<503)=P(Z<\frac{503-\mu}{\frac{\sigma}{\sqrt n}})=P(Z<\frac{503-505}{\frac{4}{\sqrt 25}})=P(Z<-2.5)=0.0062

b) P(503<xˉ<506)=P(503505425<Z<506505425)=P(2.5<Z<1.25)=P(503<\bar x<506)=P(\frac{503-505}{\frac{4}{\sqrt 25}}<Z<\frac{506-505}{\frac{4}{\sqrt 25}})=P(-2.5<Z<1.25)=

=P(Z<1.25)P(Z<2.5)=0.89440.0062=0.8882=P(Z<1.25)-P(Z<-2.5)=0.8944-0.0062=0.8882

c) P(xˉ<k)=95%=0.95    P(Z<1.645)=0.95P(\bar x<k)=95\%=0.95 \implies P(Z<1.645)=0.95

k505425=1.645    k=506.316\frac{k-505}{\frac{4}{\sqrt 25}}=1.645 \implies k=506.316


Answer:

a) 0.0062

b) 0.8882

c) 506.316


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