Question

According to an annual survey conducted by a particular website, the mean cost of a wedding in 2012 was $28,423. This is still less than the high in 2008, but it reflects increasing confidence in the economy. Suppose the cost for a wedding is normally distributed, with a standard deviation of $1,500, and a wedding is selected at random. (Round your answers to four decimal places.)(a)Find the probability that the wedding costs more than $31,000.(b)Find the probability that the wedding costs between $26,000 and $30,000.(c)Find the probability that the wedding costs less than $25,000.

Answer 1

(a) P(X > $31,000) = 0.0427

(b) P($26,000 < X < $30,000) = 0.8005

(c) P(X < $25,000) = 0.0113

Explanation:

We are given that according to an annual survey conducted by a particular website, the mean cost of a wedding in 2012 was $28,423. Suppose the cost for a wedding is normally distributed, with a standard deviation of $1,500, and a wedding is selected at random.

Let X = cost for a wedding

So, X ~ N(
`\mu=$28423,\sigma^(2) = $1,500^(2)`)

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

(a) Probability that the wedding costs more than $31,000 = P(X>$31,000)

P(X > 31,000) = P(
`(X-\mu)/(\sigma)` >
`(31,000-28,423)/(1,500)` ) = P(Z > 1.72) = 1 - P(Z <= 1.72)

= 1 - 0.95728 = 0.0427

(b) Probability that the wedding costs between $26,000 and $30,000 = P($26,000 < X < $30,000) = P(X < $30,000) - P(X <= $26,000) \

P(X < 30,000) = P(
`(X-\mu)/(\sigma)` <
`(30,000-28,423)/(1,500)` ) = P(Z < 1.05) = 0.85314

P(X <= 26,000) = P(
`(X-\mu)/(\sigma)` <=
`(26,000-28,423)/(1,500)` ) = P(Z <= -1.62) = 1 - P(Z <= 1.62)

= 1 - 0.94738 = 0.05262

Therefore, P($26,000 < X < $30,000) = 0.85314 - 0.05262 = 0.8005 .

(c) Probability that the wedding costs less than $25,000 = P(X < $25,000)

P(X < 25,000) = P(
`(X-\mu)/(\sigma)` <
`(25,000-28,423)/(1,500)` ) = P(Z < -2.28) = 1 - P(Z <= 2.28)

= 1 - 0.98870 = 0.0113

Answer 2

(a) The probability that the wedding costs more than $31,000 is 0.0427.

(b) The probability that the wedding costs between $26,000 and $30,000 is 0.8005.

(c) The probability that the wedding costs less than $25,000 is 0.0113.

Explanation:

Let X = cost of a wedding.

The random variable X follows a Normal distribution with mean, μ = $28, 423 and the standard deviation, σ = $1,500.

(a)

Compute the probability that the wedding costs more than $31,000 as follows:

`P(X>31000)=P((X-\mu)/(\sigma)>(31000-28423)/(1500))\\=P(Z>1.72)\\=1-P(Z<1.72)\\=1-0.9573\\=0.0427`

*Use a z-table for the probability.

Thus, the probability that the wedding costs more than $31,000 is 0.0427.

(b)

Compute the probability that the wedding costs between $26,000 and $30,000 as follows:

`P(26000<X<30000)=P((26000-28423)/(1500)<(X-\mu)/(\sigma)<(30000-28423)/(1500))\\=P(-1.62<Z<1.05)\\=P(Z<1.05)-P(Z<-1.62)\\=0.8531-0.0526\\=0.8005`

*Use a z-table for the probability.

Thus, the probability that the wedding costs between $26,000 and $30,000 is 0.8005.

(c)

Compute the probability that the wedding costs less than $25,000 as follows:

`P(X<25000)=P((X-\mu)/(\sigma)<(25000-28423)/(1500))\\=P(Z<-2.28)\\=1-P(Z<2.28)\\=1-0.9887\\=0.0113`

*Use a z-table for the probability.

Thus, the probability that the wedding costs less than $25,000 is 0.0113.