Question

One year consumers spent an average of $21 on a meal at a restaurant. Assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $4.

Complete parts (a) through (c) below.

a. What is the probability that a randomly selected person spent more than $ 22?

P(X> $22 )=___________________

(Round to four decimal places as needed.)

b. What is the probability that a randomly selected person spent between

$11 and $ 19?P($11<X< $19 )=_____________

(Round to four decimal places as needed.)

c. Between what two values will the middle

95 % of the amounts of cash spent fall?

The middle 95 % of the amounts of cash spent will fall between

X= $___________and X= $________________.

(Round to the nearest cent as needed.)

Answer 1

a)

$\mu=21$

$\sigma=4$

$x=22$

$Z=\frac{x-\mu}{\sigma}$

$Z=\frac{22-21}{4}$

$P(Z> \frac{22-21}{4})$

=0.401293

b)

$P(\frac{11-21}{4} < Z<\frac{19-21}{4} )$

=0.302327873 rounded answer is 0.3023

c)

x?= 21

sigma= 4

n= 1 .....(assumed as n not given )

alpha=0.05 then Z(alpha/2)=1.96

Margin of error E= Z(alpha/2)*sigma/sqrt(n)

= 1.96*4/sqrt(1)

= 7.84

95% Confidence interval for population mean =(x?-E,x?+E)

=(21-7.84,21+7.84)

lower bound= 13.16

upper bound= 28.84

$13.16 \leq \mu\leq 28.84$

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