A certain type of automobile battery is known to last an average of 1110 days with a standard deviation of 50 days. If 100 of th
1 answer:
Answer:
a. P(1104<X[bar]<1110) = 0.3849
b. P(X[bar]>1116) = 0.1151
c. P(X[bar]<940) = 0
Step-by-step explanation:
Hello!
Your study variable is X: "length of life of a battery" (days)
These batteries have a known lifetime average μ: 1110 days and a standard deviation δ:50 days.
To calculate the probabilities I'll apply the Central Limit Theorem, and since the sample is big enough (n>30), approximate the distribution of the sample mean to normal X[bar]≈N(μ;δ²/n) and use the standardization to the Z-distribution to calculate each probability.
a. The average is between 1104 and 1110
P(1104<X[bar]<1110) = P(X[bar]<1110) - P(X[bar]<1104)
= P(Z<(1110-1110)/(50/√100) - P(Z<(1104-1110)/(50/√100))
=P(Z<0) - P(Z<-1.2) = 0.5 - 0.1151 = 0.3849
b. The average is greater than 1116
P(X[bar]>1116) = 1 - P(X[bar]≤1116) = 1 - P(Z≤(1116-1110)/(50/√100))
=1 - P(Z≤1.2) = 1 - 0.8849 = 0.1151
Since the Z table accumulates from left, this means, the probability values it gives are ), to calculate the values above a certain number, you have to subtract from 1 (the highest possible cumulative probability value) all the cumulative probability that is below that number.
c. The average is less than 940.
P(X[bar]<940) = P(Z<(940-1110)/(50/√100)) = P(Z<-34) = 0
I hope you have a SUPER day!
You might be interested in