Lan pays a semiannual premium of $400 for automobile insurance, a monthly premium of $140 for health insurance, and an annual premium of $450 for life insurance. What is the monthly expense?

The monthly expense is  $244.16

Step-by-step explanation:

Given:

automobile insurance= $400

health insurance =  $140

life insurance = $450

To find:

The monthly expense =?

Solution:

The automobile insurance is Semiannual premium . so it is paid twice a year

So for a year the total automobile insurance paid is  =   =  $800

The health insurance is monthly premium. it is paid for all 12 months.

Thus the health insure for a year  is  = = $1680

The life insurance is annual premium. so it is paid once in  a year

So for a year the life insurance paid is  = $450.

The total expense for a year =  800+ 1680 + 450 = 2930

Then for one month the expense will be  = = $244.16

0.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean weight of 3215 grams, and a variance of 84,681

This means that

67 babies are sampled at random from the hospital

This means that

What is the probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams?

p-value of Z when X = 3215 + 52 = 3267 subtracted by the p-value of Z when X = 3215 - 52 = 3163. So

X = 3267

By the Central Limit Theorem

has a p-value of 0.9279

X = 3163

has a p-value of 0.0721

0.9279 - 0.0721 = 0.8558

0.8558 = 85.58% probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams.