Question

According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones, etc.) in back-to-college spending per student. Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54. If a family of a returning college student is randomly selected, what is the probability that: (a) They spend less than $150 on back-to-college electronics? (b) They spend more than $390 on back-to-college electronics? (c) They spend between $120 and $175 on back-to-college electronics?

Answer 1

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = back-to-college family spending on electronics

SO, X ~ Normal($\mu=237,\sigma^{2} =54^{2}$)

The z score probability distribution for normal distribution is given by;

                                 Z  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean family spending = $237

           $\sigma$ = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( $\frac{X-\mu}{\sigma}$ < $\frac{150-237}{54}$ ) = P(Z < -1.61) = 1 - P(Z $\leq$ 1.61)

                                                             = 1 - 0.9463 = 0.0537

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( $\frac{X-\mu}{\sigma}$ > $\frac{390-237}{54}$ ) = P(Z > 2.83) = 1 - P(Z $\leq$ 2.83)

                                                             = 1 - 0.9977 = 0.0023

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X $\leq$ $120)

     P(X < $175) = P( $\frac{X-\mu}{\sigma}$ < $\frac{175-237}{54}$ ) = P(Z < -1.15) = 1 - P(Z $\leq$ 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( $\frac{X-\mu}{\sigma}$ < $\frac{120-237}{54}$ ) = P(Z < -2.17) = 1 - P(Z $\leq$ 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = 0.1101