Question

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 these batteries are selected, find the following probabilities for the average length of life of the selected batteries. (Round your answers to four decimal places.) (a) The average is between 1104 and 1110. (b) The average is greater than 1116. (c) The average is less than 940.

Answer 1

a. P(1104<X[bar]<1110) = 0.3849

b. P(X[bar]>1116) = 0.1151

c. P(X[bar]<940) = 0

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
`P(Z<Z_(\alpha ) )`), 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!