Answer:
Step-by-step explanation:
Given that a population of 1000 students spends an average of $10.25 a day on dinner.
The standard deviation of the expenditure is $3.
X , the amount spend is N(10.25, 3)
n =64
a) Mean of sample mean = 10.25
Std devition of sample mean = std deviation /sq rt of n = 0.375
The shape would be bell shaped symmetrical about the mean and unimodal.
b) Prob that these 64 students will spend a combined total of more than $715.21
= [tex]P(\bar x >\frac{715.21}{64} )\\\\=P(\bar x >11.1752)\\=1-0.9932\\=0.0068[/tex]
c) the probability that these 64 students will spend a combined total between $703.59 and $728.45
= x bar lies between = 10.9936 and 11.38203
=[tex]F(11.38203)-F(10.9936)\\\=0.9987-0.9763\\=0.0224[/tex]
